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Solve the following kinematics problems involving displacement, initial velocity, acceleration, and time.

Physics

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Part 31· 3 steps

To find the time taken, we use the kinematic equation for displacement.
Given:
Initial velocity u=0 m/su = 0 \text{ m/s} (starts from rest)
Displacement s=98 ms = 98 \text{ m}
Acceleration a=4 m/s2a = 4 \text{ m/s}^2

  1. Use the kinematic equation s=ut+12at2s = ut + \frac{1}{2}at^2.

    98=(0)t+12(4)t298 = (0)t + \frac{1}{2}(4)t^2

  2. Simplify the equation.

    98=2t298 = 2t^2

  3. Solve for tt.

    t2=982t^2 = \frac{98}{2} t2=49t^2 = 49 t=49t = \sqrt{49} t=7 st = 7 \text{ s}

    The correct option is A.

    The time taken is A. 7 sec\boxed{\text{A. 7 sec}}.

Part 32· 5 steps

To find the time taken, we use the kinematic equation for displacement.
Given:
Initial velocity u=72 km/hu = 72 \text{ km/h}
Deceleration a=−1.5 m/s2a = -1.5 \text{ m/s}^2
Displacement s=70 ms = 70 \text{ m}

  1. Convert initial velocity from km/h to m/s.

    u=72 km/h×1000 m1 km×1 h3600 su = 72 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} u=20 m/su = 20 \text{ m/s}

  2. Use the kinematic equation s=ut+12at2s = ut + \frac{1}{2}at^2.

    70=(20)t+12(−1.5)t270 = (20)t + \frac{1}{2}(-1.5)t^2 70=20t−0.75t270 = 20t - 0.75t^2

  3. Rearrange into a quadratic equation At2+Bt+C=0At^2 + Bt + C = 0.

    0.75t2−20t+70=00.75t^2 - 20t + 70 = 0

  4. Use the quadratic formula t=−B±B2−4AC2At = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}.

    Here, A=0.75A = 0.75, B=−20B = -20, C=70C = 70.

    t=−(−20)±(−20)2−4(0.75)(70)2(0.75)t = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(0.75)(70)}}{2(0.75)} t=20±400−2101.5t = \frac{20 \pm \sqrt{400 - 210}}{1.5} t=20±1901.5t = \frac{20 \pm \sqrt{190}}{1.5} t=20±13.7841.5t = \frac{20 \pm 13.784}{1.5}

  5. Calculate the two possible values for tt.

    t1=20+13.7841.5=33.7841.5≈22.52 st_1 = \frac{20 + 13.784}{1.5} = \frac{33.784}{1.5} \approx 22.52 \text{ s} t2=20−13.7841.5=6.2161.5≈4.144 st_2 = \frac{20 - 13.784}{1.5} = \frac{6.216}{1.5} \approx 4.144 \text{ s}

    Since the car is slowing down, both times are physically possible if the car hasn't stopped and reversed within 70m. However, usually, the smaller positive time is the one sought in such problems before it potentially stops or reverses. Let's check the final velocity for t2t_2.
    v=u+at=20+(−1.5)(4.144)=20−6.216=13.784 m/sv = u + at = 20 + (-1.5)(4.144) = 20 - 6.216 = 13.784 \text{ m/s}. This is a valid positive velocity.
    The correct option is C.

    The time taken is C. 4.14 sec\boxed{\text{C. 4.14 sec}}.

Part 33

The phenomenon where a ball strikes the ground (action) and the ground pushes back with an equal and opposite force (reaction) is a direct application of Newton's Third Law of Motion.

The correct option is C.

The law of motion is C. third law\boxed{\text{C. third law}}.

Part 34

The statement "the vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration a of the object" is the definition of Newton's Second Law of Motion, often expressed as F=maF = ma.

The correct option is B.

The name of this law of motion is B. second law\boxed{\text{B. second law}}.

Part 35

When you touch something, you exert a force on it, and in response, the object exerts an equal and opposite force back on you. It is this reaction force that you feel. This interaction is explained by Newton's Third Law of Motion.

The correct option is C.

The law is C. third law\boxed{\text{C. third law}}.

Part 36· 2 steps

Work done is calculated as the product of force and displacement in the direction of the force.
Given:
Force F=10 NF = 10 \text{ N}
Displacement d=4 md = 4 \text{ m}

  1. Use the formula for work done W=F×dW = F \times d.

    W=10 N×4 mW = 10 \text{ N} \times 4 \text{ m}

  2. Calculate the work done.

    W=40 JW = 40 \text{ J}

    The correct option is A.

    The work done is A. 40 J\boxed{\text{A. 40 J}}.

Part 37· 2 steps

Work done to lift an object is equal to the force (weight) multiplied by the height lifted.
Given:
Work done W=500 JW = 500 \text{ J}
Force (weight) F=200 NF = 200 \text{ N}

  1. Use the formula for work done W=F×hW = F \times h.

    500 J=200 N×h500 \text{ J} = 200 \text{ N} \times h

  2. Solve for hh.

    h=500 J200 Nh = \frac{500 \text{ J}}{200 \text{ N}} h=2.5 mh = 2.5 \text{ m}

    The correct option is B.

    The height the weight was lifted is B. 2.5 m\boxed{\text{B. 2.5 m}}.

Part 38· 2 steps

Power is the rate at which energy is transformed or transferred.
Given:
Energy E=10,000 JE = 10,000 \text{ J}
Time t=4 st = 4 \text{ s}

  1. Use the formula for power P=EtP = \frac{E}{t}.

    P=10,000 J4 sP = \frac{10,000 \text{ J}}{4 \text{ s}}

  2. Calculate the power.

    P=2500 WP = 2500 \text{ W}

    The correct option is B.

    The power is B. 2,500 W\boxed{\text{B. 2,500 W}}.

Part 39· 2 steps

Energy transformed is the product of power and time.
Given:
Power P=60 WP = 60 \text{ W}
Time t=20 st = 20 \text{ s}

  1. Use the formula for energy E=P×tE = P \times t.

    E=60 W×20 sE = 60 \text{ W} \times 20 \text{ s}

  2. Calculate the energy transformed.

    E=1200 JE = 1200 \text{ J}

    The correct option is A.

    The energy transformed is A. 1,200 J\boxed{\text{A. 1,200 J}}.

Part 40

Kinetic energy is the energy an object possesses due to its motion.

  • A body at rest has zero kinetic energy.
  • A body in motion possesses kinetic energy.
  • Being placed on a platform does not inherently mean motion.
  • Zero gravity does not prevent motion or kinetic energy.

The correct option is B.

A body possesses kinetic energy in the state of B. Motion\boxed{\text{B. Motion}}.

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