QUESTIONS FROM MOTION I already sent 1-10 to the telegram class group. I'll explain some and the others are for you to practice. A body starts from rest and moves with a uniform acceleration of 5 m/s2 for 10 s. It then decelerates uniformly to rest in another 5 s. Calculate the total distance covered. A) 250 m B) 375 m C) 500 m D) 625 m The displacement x of a particle varies with time t as x = 2t3 - 6t2 + 4. At what time t will the acceleration of the particle become zero? A) 1 s B) 2 s C) 3 s D) 4 s A bullet is fired vertically upwards from a point on the ground with an initial speed of 80 m/s. Find its height above the ground after 10 s. [g = 10 m/s2] A) 300 m B) 400 m C) 500 m D) 800 m A car moves around a circular path of radius 14 m with a constant speed of 22 m/s. Calculate its centripetal acceleration. A) 1.57 m/s2 B) 22.0 m/s2 C) 34.57 m/s2 D) 44.0 m/s2 A ball is dropped from a cliff 80 m high. Concurrently, another ball is projected vertically downwards from the same height with an initial velocity of 10 m/s. What is the difference in their times of flight to hit the ground? [g = 10 m/s2] A) 0.5 s B) 1.0 s C) 2.0 s D) 3.0 s The velocity-time graph of a moving body is a straight line inclined to the time axis at an angle θ. The slope of this line represents the body's: A) Distance B) Displacement C) Acceleration D) Velocity A particle's position is described by s(t) = 4t2 - 3t + 5. Calculate its instantaneous velocity at t = 3 s. A) 12 m/s B) 21 m/s C) 24 m/s D) 32 m/s An object is dropped from a high balloon rising vertically at a constant speed of 5 m/s. If the balloon is at a height of 100 m when the object is released, how long does the object take to reach the ground? [g = 10 m/s2] A) 4.0 s B) 4.5 s C) 5.0 s D) 5.5 s A body covers a distance of 20 m in the 2nd second and 30 m in the 4th second of its uniformly accelerated path. Find its initial velocity u. A) 10 m/s B) 12.5 m/s C) 15 m/s D) 17.5 m/s A stone is dropped down a deep well, and the sound of the splash is heard 3.2 s later. If the speed of sound is 340 m/s, calculate the approximate depth of the well. [g = 10 m/s2] A) 45 m B) 51 m C) 64 m D) 80 m A vehicle travels the first half of a straight distance at a |uniform speed of 30 km/h and the second half of the distance at 60 km/h. The average speed of the vehicle for the entire journey is: A) 40 km/h B) 45 km/h C) 50 km/h D) 55 km/h A stone is thrown horizontally from the top of a tower 45 m high with a velocity of 15 m/s. Find the horizontal distance from the foot of the tower where the stone hits the ground. [g = 10 m/s2] A) 30 m B) 45 m C) 60 m D) 75 m The acceleration a of a body moving in a straight line is given by a = 6t. If the body starts from rest, find the distance traveled after 2 s. A) 4 m B) 6 m C) 8 m D) 12 m Two bodies of masses 2 kg and 10 kg are dropped simultaneously from the same height in a vacuum. Which body reaches the ground first? A) The 10 kg body B) The 2 kg body C) Both reach the ground at the same time D) It depends on atmospheric pressure A river is flowing from west to east at a speed of 5 m/min. A man on the south bank capable of swimming at 10 m/min in still water wants to swim across the river along the shortest path. In what direction should he swim? A) Due North B) 30° West of North C) 30° East of North D) 60° West of North A driver traveling at 20 m/s sees a hazard and applies the brakes, bringing the car to a halt under uniform deceleration over a distance of 40 m. What was the reaction time of the driver if the total distance from hazard visibility to full stop is 45 m? A) 0.25 s B) 0.50 s /C) 0.75 s D) 1.00 s A displacement-time graph consists of a curve whose slope decreases progressively until it becomes horizontal. This profile indicates that the object is: A) Accelerating uniformly B) Moving with constant speed C) Decelerating to rest D) Undergoing periodic motion Find the angle of projection for a projectile whose maximum height reached is exactly equal to its horizontal range. A) θ = tan^-1(1) B) θ = tan^-1(2) C) θ = tan^-1(4) D) θ = tan^-1(0.25) A body moves with a velocity profile represented by v = 4t - t2. Determine the maximum velocity reached by the body. A) 2 m/s B) 4 m/s C) 6 m/s D) 8 m/s A tape attached to a moving trolley passes through a ticker-timer operating at a frequency of 50 Hz. If the distance between 5 consecutive dots is 10 cm, calculate the average speed of the trolley over this interval. A) 0.5 m/s B) 1.25 m/s C) 2.0 m/s D) 2.5 m/s A stone tied to a string of length 0.5 m is whirled in a horizontal circle. If it completes 4 revolutions every second, find its linear speed. A) 2π m/s B) 4π m/s C) 8π m/s D) 16π m/s A vector quantities representation shows displacement components along the x and y axes to be 6 m and 8 m respectively. What is the resultant magnitude of this displacement? A) 10 m B) 14 m C) 48 m D) 100 m A particle executes uniform circular motion. Which of the following quantities remains constant throughout the motion? A) Velocity B) Acceleration C) Speed D) Displacement A rocket is launched vertically upwards with a constant net acceleration of 5 m/s2. After 10 s, its fuel is completely exhausted. To what maximum height will the rocket rise from the ground? [g = 10 m/s2] A) 250 m B) 375 m C) 500 m D) 625 m A sports car travels down a track according to the function s = t3 - 3t. Find its acceleration at the instant its velocity is exactly zero. A) 0 m/s2 B) 3 m/s2 C) 6 m/s2 D) 9 m/s2

1. A body starts from rest and moves with a uniform acceleration of 5 m/s^2 for 10 s. It then decelerates uniformly to rest in another 5 s. Calculate the total distance covered. A) 250 m B) 375 m C) 500 m D) 625 m

Step 1: Calculate the distance covered during acceleration. Initial velocity $u_1 = 0$ m/s, acceleration $a_1 = 5$ m/s$^2$, time $t_1 = 10$ s. The distance $s_1$ is given by: $s_1 = u_1 t_1 + \frac{1}{2} a_1 t_1^2$ $s_1 = (0)(10) + \frac{1}{2}(5)(10)^2$ $s_1 = 0 + \frac{1}{2}(5)(100)$ $s_1 = 250 \text{ m}$

Step 2: Calculate the velocity at the end of the acceleration phase. This velocity becomes the initial velocity for the deceleration phase. $v_1 = u_1 + a_1 t_1$ $v_1 = 0 + (5)(10)$ $v_1 = 50 \text{ m/s}$

Step 3: Calculate the distance covered during deceleration. Initial velocity $u_2 = v_1 = 50$ m/s, final velocity $v_2 = 0$ m/s, time $t_2 = 5$ s. The distance $s_2$ is given by: $s_2 = \left(\frac{u_2 + v_2}{2}\right) t_2$ $s_2 = \left(\frac{50 + 0}{2}\right) (5)$ $s_2 = (25)(5)$ $s_2 = 125 \text{ m}$

Step 4: Calculate the total distance covered. $S_{total} = s_1 + s_2$ $S_{total} = 250 m + 125 \text{ m}$ $S_{total} = 375 \text{ m}$ The correct option is B. $\boxed{B) 375 m}$ Send me the next one 📸

2. The displacement x of a particle varies with time t as x = 2t^3 - 6t^2 + 4. At what time t will the acceleration of the particle become zero? A) 1 s B) 2 s C) 3 s D) 4 s

Step 1: Find the velocity function by differentiating the displacement function with respect to time. Given displacement $x(t) = 2t^3 - 6t^2 + 4$. Velocity $v(t) = \frac{dx}{dt}$: $v(t) = \frac{d}{dt}(2t^3 - 6t^2 + 4)$ $v(t) = 6t^2 - 12t$

Step 2: Find the acceleration function by differentiating the velocity function with respect to time. Acceleration $a(t) = \frac{dv}{dt}$: $a(t) = \frac{d}{dt}(6t^2 - 12t)$ $a(t) = 12t - 12$

Step 3: Set the acceleration to zero and solve for time $t$. $a(t) = 0$ $12t - 12 = 0$ $12t = 12$ $t = 1 \text{ s}$ The correct option is A. $\boxed{A) 1 s}$ What's next?

3. A bullet is fired vertically upwards from a point on the ground with an initial speed of 80 m/s. Find its height above the ground after 10 s. [g = 10 m/s^2] A) 300 m B) 400 m C) 500 m D) 800 m

Step 1: Identify the given values and the appropriate kinematic equation. Initial velocity $u = 80$ m/s (upwards). Acceleration due to gravity $a = -g = -10$ m/s$^2$ (downwards). Time $t = 10$ s. We need to find the height $h$. The equation of motion is: $h = ut + \frac{1}{2}at^2$

Step 2: Substitute the values into the equation and calculate the height. $h = (80)(10) + \frac{1}{2}(-10)(10)^2$ $h = 800 - \frac{1}{2}(10)(100)$ $h = 800 - 500$ $h = 300 \text{ m}$ The correct option is A. $\boxed{A) 300 m}$ Got more? Send 'em!

4. A car moves around a circular path of radius 14 m with a constant speed of 22 m/s. Calculate its centripetal acceleration. A) 1.57 m/s^2 B) 22.0 m/s^2 C) 34.57 m/s^2 D) 44.0 m/s^2

Step 1: Identify the given values. Radius of the circular path $r = 14$ m. Constant speed $v = 22$ m/s.

Step 2: Use the formula for centripetal acceleration. The centripetal acceleration $a_c$ is given by: $a_c = \frac{v^2}{r}$

Step 3: Substitute the values and calculate the centripetal acceleration. $a_c = \frac{(22 m/s)^2}{14 m}$ $a_c = \frac{484 m^2/s^2}{14 m}$ $a_c \approx 34.5714 m/s^2$ Rounding to two decimal places, $a_c \approx 34.57$ m/s$^2$. The correct option is C. $\boxed{C) 34.57 m/s^2}$ Drop the next question.

5. A ball is dropped from a cliff 80 m high. Concurrently, another ball is projected vertically downwards from the same height with an initial velocity of 10 m/s. What is the difference in their times of flight to hit the ground? [g = 10 m/s^2] A) 0.5 s B) 1.0 s C) 2.0 s D) 3.0 s

Step 1: Calculate the time of flight for the first ball (dropped). Height $h = 80$ m, initial velocity $u_1 = 0$ m/s, acceleration $a = g = 10$ m/s$^2$. Using the equation $h = u_1 t_1 + \frac{1}{2}gt_1^2$: $80 = (0)t_1 + \frac{1}{2}(10)t_1^2$ $80 = 5t_1^2$ $t_1^2 = \frac{80}{5}$ $t_1^2 = 16$ $t_1 = \sqrt{16} = 4 \text{ s}$

Step 2: Calculate the time of flight for the second ball (projected downwards). Height $h = 80$ m, initial velocity $u_2 = 10$ m/s, acceleration $a = g = 10$ m/s$^2$. Using the equation $h = u_2 t_2 + \frac{1}{2}gt_2^2$: $80 = 10t_2 + \frac{1}{2}(10)t_2^2$ $80 = 10t_2 + 5t_2^2$ Rearrange into a quadratic equation: $5t_2^2 + 10t_2 - 80 = 0$ Divide by 5: $t_2^2 + 2t_2 - 16 = 0$

Step 3: Solve the quadratic equation for $t_2$. Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $t_2 = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-16)}}{2(1)}$ $t_2 = \frac{-2 \pm \sqrt{4 + 64}}{2}$ $t_2 = \frac{-2 \pm \sqrt{68}}{2}$ $t_2 = \frac{-2 \pm 8.246}{2}$ Since time cannot be negative, we take the positive root: $t_2 = \frac{-2 + 8.246}{2}$ $t_2 = \frac{6.246}{2}$ $t_2 \approx 3.123 \text{ s}$

Step 4: Calculate the difference in their times of flight. $\Delta t = t_1 - t_2$ $\Delta t = 4 s - 3.123 \text{ s}$ $\Delta t \approx 0.877 \text{ s}$ This is closest to 1.0 s. Let's recheck the calculation for $t_2$. If we use $g=9.8$ m/s$^2$, the values would be slightly different. Given $g=10$ m/s$^2$, let's check the options. If $\Delta t = 1.0$ s, then $t_2 = 3.0$ s. Let's check if $t_2 = 3.0$ s satisfies $5t_2^2 + 10t_2 - 80 = 0$: $5(3)^2 + 10(3) - 80 = 5(9) + 30 - 80 = 45 + 30 - 80 = 75 - 80 = -5$. This is close to 0. Let's re-evaluate the quadratic formula. $t_2^2 + 2t_2 - 16 = 0$. If $t_2 = 3$ s, $9+6-16 = -1$. If $t_2 = 2.8$ s, $(2.8)^2 + 2(2.8) - 16 = 7.84 + 5.6 - 16 = 13.44 - 16 = -2.56$. If $t_2 = 3.2$ s, $(3.2)^2 + 2(3.2) - 16 = 10.24 + 6.4 - 16 = 16.64 - 16 = 0.64$. The value $t_2 \approx 3.123$ s is correct. The difference is $4 - 3.123 = 0.877$ s. Among the given options, 1.0 s is the closest. It's possible the question expects an approximation or there's a slight discrepancy in the options. Given the options, 1.0 s is the most reasonable choice.

Let's consider if there's a simpler way or if I made a mistake. $t_1 = 4$ s. For $t_2$: $5t_2^2 + 10t_2 - 80 = 0 \implies t_2^2 + 2t_2 - 16 = 0$. The roots are $t_2 = \frac{-2 \pm \sqrt{4 - 4(1)(-16)}}{2} = \frac{-2 \pm \sqrt{4+64}}{2} = \frac{-2 \pm \sqrt{68}}{2}$. $\sqrt{68} \approx 8.246$. $t_2 = \frac{-2 + 8.246}{2} = \frac{6.246}{2} = 3.123$ s. Difference $= 4 - 3.123 = 0.877$ s. This is closest to 1.0 s.

Let's assume the options are exact and check if any of them lead to a simpler solution or if there's an error in my calculation or interpretation. If the difference is 1.0 s, then $t_2 = 3.0$ s. If $t_2 = 3.0$ s, then $h = 10(3) + \frac{1}{2}(10)(3)^2 = 30 + 5(9) = 30 + 45 = 75$ m. This is not 80 m. So, $t_2$ is not exactly 3.0 s. However, in multiple-choice questions, sometimes the closest answer is expected. The difference is $0.877$ s, which is closer to $1.0$ s than to $0.5$ s or $2.0$ s. So, I will select B. $\boxed{B) 1.0 s}$ Send me the next one!

6. The velocity-time graph of a moving body is a straight line inclined to the time axis at an angle θ. The slope of this line represents the body's: A) Distance B) Displacement C) Acceleration D) Velocity

Step 1: Recall the definition of the slope of a velocity-time graph. The slope of any graph is given by the change in the vertical axis quantity divided by the change in the horizontal axis quantity. For a velocity-time graph, the vertical axis is velocity ($v$) and the horizontal axis is time ($t$).

Step 2: Determine what the ratio $\frac{\Delta v}{\Delta t}$ represents. The change in velocity over the change in time ($\frac{\Delta v}{\Delta t}$) is the definition of acceleration. Therefore, the slope of a velocity-time graph represents the body's acceleration. The correct option is C. $\boxed{C) Acceleration}$ What's next?

7. A particle's position is described by s(t) = 4t^2 - 3t + 5. Calculate its instantaneous velocity at t = 3 s. A) 12 m/s B) 21 m/s C) 24 m/s D) 32 m/s

Step 1: Find the instantaneous velocity function by differentiating the position function with respect to time. Given position $s(t) = 4t^2 - 3t + 5$. Instantaneous velocity $v(t) = \frac{ds}{dt}$: $v(t) = \frac{d}{dt}(4t^2 - 3t + 5)$ $v(t) = 8t - 3$

Step 2: Substitute $t = 3$ s into the velocity function to find the instantaneous velocity at that time. $v(3) = 8(3) - 3$ $v(3) = 24 - 3$ $v(3) = 21 \text{ m/s}$ The correct option is B. $\boxed{B) 21 m/s}$ Got more? Send 'em!

8. An object is dropped from a high balloon rising vertically at a constant speed of 5 m/s. If the balloon is at a height of 100 m when the object is released, how long does the object take to reach the ground? [g = 10 m/s^2] A) 4.0 s B) 4.5 s C) 5.0 s D) 5.5 s

Step 1: Identify the initial conditions for the object. When the object is released from the rising balloon, it initially has the same upward velocity as the balloon. Initial velocity $u = +5$ m/s (upwards). Displacement $s = -100$ m (since it falls 100 m downwards from its release point). Acceleration $a = -g = -10$ m/s$^2$ (due to gravity, acting downwards).

Step 2: Use the kinematic equation relating displacement, initial velocity, acceleration, and time. $s = ut + \frac{1}{2}at^2$ $-100 = (5)t + \frac{1}{2}(-10)t^2$ $-100 = 5t - 5t^2$

Step 3: Rearrange the equation into a standard quadratic form and solve for $t$. $5t^2 - 5t - 100 = 0$ Divide the entire equation by 5: $t^2 - t - 20 = 0$ Factor the quadratic equation: $(t - 5)(t + 4) = 0$ This gives two possible solutions for $t$: $t = 5$ s or $t = -4$ s. Since time cannot be negative, we take the positive value. $t = 5 \text{ s}$ The correct option is C. $\boxed{C) 5.0 s}$ Drop the next question.

9. A body covers a distance of 20 m in the 2nd second and 30 m in the 4th second of its uniformly accelerated path. Find its initial velocity u. A) 10 m/s B) 12.5 m/s C) 15 m/s D) 17.5 m/s

Step 1: Use the formula for distance covered in the $n$-th second. The distance covered in the $n$-th second is given by: $s_n = u + \frac{a}{2}(2n - 1)$ where $u$ is the initial velocity and $a$ is the uniform acceleration.

Step 2: Set up equations for the given information. For the 2nd second ($n=2$), $s_2 = 20$ m: $20 = u + \frac{a}{2}(2(2) - 1)$ $20 = u + \frac{3a}{2} \quad \text{(Equation 1)}$ For the 4th second ($n=4$), $s_4 = 30$ m: $30 = u + \frac{a}{2}(2(4) - 1)$ $30 = u + \frac{7a}{2} \quad \text{(Equation 2)}$

Step 3: Solve the system of two linear equations for $u$ and $a$. Subtract Equation 1 from Equation 2: $(30 - 20) = \left(u + \frac{7a}{2}\right) - \left(u + \frac{3a}{2}\right)$ $10 = \frac{7a}{2} - \frac{3a}{2}$ $10 = \frac{4a}{2}$ $10 = 2a$ $a = 5 m/s^2$

Step 4: Substitute the value of $a$ back into Equation 1 to find $u$. $20 = u + \frac{3(5)}{2}$ $20 = u + \frac{15}{2}$ $20 = u + 7.5$ $u = 20 - 7.5$ $u = 12.5 \text{ m/s}$ The correct option is B. $\boxed{B) 12.5 m/s}$ Send me the next one 📸

10. A stone is dropped down a deep well, and the sound of the splash is heard 3.2 s later. If the speed of sound is 340 m/s, calculate the approximate depth of the well. [g = 10 m/s^2] A) 45 m B) 51 m C) 64 m D) 80 m

Step 1: Define variables and set up equations for the time taken for the stone to fall and the sound to travel up. Let $h$ be the depth of the well. Time for the stone to fall ($t_f$): The stone is dropped, so initial velocity $u=0$. Using $h = ut_f + \frac{1}{2}gt_f^2$: $h = 0 \cdot t_f + \frac{1}{2}(10)t_f^2$ $h = 5t_f^2 \implies t_f = \sqrt{\frac{h}{5}}$ Time for the sound to travel up ($t_s$): Speed of sound $v_s = 340$ m/s. $h = v_s t_s \implies t_s = \frac{h}{340}$

Step 2: Use the total time given to form an equation. Total time $T = t_f + t_s = 3.2$ s. $3.2 = \sqrt{\frac{h}{5}} + \frac{h}{340}$

Step 3: Solve for $h$. This equation is difficult to solve directly. We can test the given options. Let's test option A, $h = 45$ m: $t_f = \sqrt{\frac{45}{5}} = \sqrt{9} = 3 \text{ s}$ $t_s = \frac{45}{340} \approx 0.132 \text{ s}$ $T = t_f + t_s = 3 + 0.132 = 3.132 \text{ s}$ This value (3.132 s) is very close to the given total time of 3.2 s.

Let's quickly check other options to confirm. For $h = 51$ m: $t_f = \sqrt{51/5} = \sqrt{10.2} \approx 3.19$ s. $t_s = 51/340 \approx 0.15$ s. $T \approx 3.34$ s. (Too high) For $h = 64$ m: $t_f = \sqrt{64/5} = \sqrt{12.8} \approx 3.58$ s. $t_s = 64/340 \approx 0.19$ s. $T \approx 3.77$ s. (Too high) Option A gives the closest approximation. The correct option is A. $\boxed{A) 45 m}$ What's next?

11. A vehicle travels the first half of a straight distance at a uniform speed of 30 km/h and the second half of the distance at 60 km/h. The average speed of the vehicle for the entire journey is: A) 40 km/h B) 45 km/h C) 50 km/h D) 55 km/h

Step 1: Define variables for the distance and times. Let the total distance be $2D$. The first half of the distance is $D$, covered at speed $v_1 = 30$ km/h. The second half of the distance is $D$, covered at speed $v_2 = 60$ km/h.

Step 2: Calculate the time taken for each half of the journey. Time for the first half $t_1 = \frac{Distance}{Speed} = \frac{D}{v_1} = \frac{D}{30}$. Time for the second half $t_2 = \frac{Distance}{Speed} = \frac{D}{v_2} = \frac{D}{60}$.

Step 3: Calculate the total distance and total time. Total distance $= D + D = 2D$. Total time $= t_1 + t_2 = \frac{D}{30} + \frac{D}{60}$. To add the fractions, find a common denominator (60): Total time $= \frac{2D}{60} + \frac{D}{60} = \frac{3D}{60} = \frac{D}{20}$.

Step 4: Calculate the average speed. Average speed $= \frac{Total distance}{Total time}$ $\text{Average speed} = \frac{2D}{\frac{D}{20}}$ $\text{Average speed} = 2D \times \frac{20}{D}$ $\text{Average speed} = 40 \text{ km/h}$ The correct option is A. $\boxed{A) 40 km/h}$ Got more? Send 'em!

12. A stone is thrown horizontally from the top of a tower 45 m high with a velocity of 15 m/s. Find the horizontal distance from the foot of the tower where the stone hits the ground. [g = 10 m/s^2] A) 30 m B) 45 m C) 60 m D) 75 m

Step 1: Analyze the vertical motion to find the time of flight. Height of the tower $h = 45$ m. Initial vertical velocity $u_y = 0$ m/s (since it's thrown horizontally). Acceleration due to gravity $a_y = g = 10$ m/s$^2$. Using the kinematic equation $h = u_y t + \frac{1}{2}a_y t^2$: $45 = (0)t + \frac{1}{2}(10)t^2$ $45 = 5t^2$ $t^2 = \frac{45}{5}$ $t^2 = 9$ $t = \sqrt{9} = 3 \text{ s}$ The time of flight is 3 seconds.

Step 2: Analyze the horizontal motion to find the horizontal distance. Initial horizontal velocity $u_x = 15$ m/s (constant, as there is no horizontal acceleration). Time of flight $t = 3$ s. Horizontal distance $R = u_x t$: $R = (15 m/s)(3 s)$ $R = 45 \text{ m}$ The correct option is B. $\boxed{B) 45 m}$ Drop the next question.

13. The acceleration a of a body moving in a straight line is given by a = 6t. If the body starts from rest, find the distance traveled after 2 s. A) 4 m B) 6 m C) 8 m D) 12 m

Step 1: Find the velocity function by integrating the acceleration function. Given acceleration $a(t) = 6t$. Velocity $v(t) = \int a(t)\,dt = \int 6t\,dt$. $v(t) = 3t^2 + C_1$ Since the body starts from rest, $v(0) = 0$. $0 = 3(0)^2 + C_1 \implies C_1 = 0$ So, the velocity function is $v(t) = 3t^2$.

Step 2: Find the distance function by integrating the velocity function. Distance $s(t) = \int v(t)\,dt = \int 3t^2\,dt$. $s(t) = t^3 + C_2$ Assuming the body starts at position $s(0) = 0$. $0 = (0)^3 + C_2 \implies C_2 = 0$ So, the distance function is $s(t) = t^3$.

Step 3: Calculate the distance traveled after $t = 2$ s. $s(2) = (2)^3$ $s(2) = 8 \text{ m}$ The correct option is C. $\boxed{C) 8 m}$ Send me the next one 📸

14. Two bodies of masses 2 kg and 10 kg are dropped simultaneously from the same height in a vacuum. Which body reaches the ground first? A) The 10 kg body B) The 2 kg body C) Both reach the ground at the same time D) It depends on atmospheric pressure

Step 1: Understand the concept of free fall in a vacuum. In a vacuum, there is no air resistance. The acceleration due to gravity ($g$) is constant for all objects, regardless of their mass.

Step 2: Apply the concept to the given scenario. Since both bodies are dropped from the same height and experience the same acceleration ($g$) in a vacuum, their motion will be identical. They will fall at the same rate. Therefore, both bodies will reach the ground at the same time. The correct option is C. $\boxed{C) Both reach the ground at the same time}$ What's next?

15. A river is flowing from west to east at a speed of 5 m/min. A man on the south bank capable of swimming at 10 m/min in still water wants to swim across the river along the shortest path. In what direction should he swim? A) Due North B) 30° West of North C) 30° East of North D) 60° West of North

Step 1: Understand the goal: shortest path across the river. To swim across the river along the shortest path, the man's resultant velocity relative to the ground must be directly perpendicular to the river flow, i.e., due North.

Step 2: Analyze the velocities involved. Velocity of the river $v_r = 5$ m/min (East). Velocity of the man relative to still water $v_s = 10$ m/min. Let the man swim at an angle $\theta$ West of North relative to the water. The components of the man's velocity relative to the water are: • Northward component: $v_s \cos\theta$ • Westward component: $v_s \sin\theta$

Step 3: Set up the condition for the resultant velocity to be due North. For the resultant velocity to be purely North, the man's westward component of velocity must exactly cancel the river's eastward velocity. $v_s \sin\theta = v_r$ $10 \sin\theta = 5$ $\sin\theta = \frac{5}{10}$ $\sin\theta = \frac{1}{2}$

Step 4: Calculate the angle $\theta$. $\theta = \arcsin\left(\frac{1}{2}\right)$ $\theta = 30^\circ$ So, the man should swim at an angle of 30° West of North. The correct option is B. $\boxed{B) 30° West of North}$ Got more? Send 'em!

16. A driver traveling at 20 m/s sees a hazard and applies the brakes, bringing the car to a halt under uniform deceleration over a distance of 40 m. What was the reaction time of the driver if the total distance from hazard visibility to full stop is 45 m? A) 0.25 s B) 0.50 s C) 0.75 s D) 1.00 s

Step 1: Determine the distance covered during the reaction time. Total distance from hazard visibility to full stop = 45 m. Braking distance (distance covered while decelerating) = 40 m. Distance covered during reaction time $s_{reaction} = Total distance - \text{Braking distance}$. $s_{reaction} = 45 m - 40 \text{ m}$ $s_{reaction} = 5 \text{ m}$

Step 2: Calculate the reaction time. During the reaction time, the car continues to move at its initial constant speed before braking. Initial speed $u = 20$ m/s. Reaction time $t_{reaction} = \frac{s_{reaction}}{u}$. $t_{reaction} = \frac{5 m}{20 m/s}$ $t_{reaction} = 0.25 \text{ s}$ The correct option is A. $\boxed{A) 0.25 s}$ Drop the next question.

17. A displacement-time graph consists of a curve whose slope decreases progressively until it becomes horizontal. This profile indicates that the object is: A) Accelerating uniformly B) Moving with constant speed C) Decelerating to rest D) Undergoing periodic motion

Step 1: Understand what the slope of a displacement-time graph represents. The slope of a displacement-time graph represents the velocity of the object.

Step 2: Interpret the given characteristics of the slope. • "Slope decreases progressively": This means the velocity of the object is decreasing. • "until it becomes horizontal": A horizontal line on a displacement-time graph has a slope of zero, which means the velocity is zero. Combining these, the object's velocity is decreasing until it stops. This describes deceleration to rest. The correct option is C. $\boxed{C) \textbf{Decelerating to rest

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