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A car accelerates from rest to 20 m/s in 10 s, maintains this speed for 20 s, and then decelerates to rest in 5 s. Calculate the total distance traveled.

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Answers

2 parts
i

550 m\text{550 m}

ii

15.71 m/s\text{15.71 m/s}

4.04 s\text{4.04 s}

Step-by-step solution

9 steps

Here's the solution to the problems from the image you sent:

: Car Motion

Based on the description, I will assume the following values for the car's motion:

  • The car accelerates from rest to a velocity of 20 m/s20 \text{ m/s} in 10 s10 \text{ s}.
  • It then maintains this constant velocity for 20 s20 \text{ s}.
  • Finally, it decelerates to rest in 5 s5 \text{ s}.

Part i· 4 steps

Calculate the total distance traveled by the car

  1. Calculate the distance traveled during the acceleration phase (Phase 1).

    This is the area of a triangle on a velocity-time graph.

    d1=12×base×heightd_1 = \frac{1}{2} \times \text{base} \times \text{height} d1=12×10 s×20 m/sd_1 = \frac{1}{2} \times 10 \text{ s} \times 20 \text{ m/s} d1=100 md_1 = 100 \text{ m}

  2. Calculate the distance traveled during the constant velocity phase (Phase 2).

    This is the area of a rectangle.

    d2=velocity×timed_2 = \text{velocity} \times \text{time} d2=20 m/s×20 sd_2 = 20 \text{ m/s} \times 20 \text{ s} d2=400 md_2 = 400 \text{ m}

  3. Calculate the distance traveled during the deceleration phase (Phase 3).

    This is the area of a triangle.

    d3=12×base×heightd_3 = \frac{1}{2} \times \text{base} \times \text{height} d3=12×5 s×20 m/sd_3 = \frac{1}{2} \times 5 \text{ s} \times 20 \text{ m/s} d3=50 md_3 = 50 \text{ m}

  4. Calculate the total distance traveled.

    Dtotal=d1+d2+d3D_{\text{total}} = d_1 + d_2 + d_3 Dtotal=100 m+400 m+50 mD_{\text{total}} = 100 \text{ m} + 400 \text{ m} + 50 \text{ m} Dtotal=550 mD_{\text{total}} = \boxed{\text{550 m}}

Part ii

15.71 m/s\text{15.71 m/s}

What is the average speed of the car?

  1. Calculate the total time taken.

    Ttotal=t1+t2+t3T_{\text{total}} = t_1 + t_2 + t_3 Ttotal=10 s+20 s+5 sT_{\text{total}} = 10 \text{ s} + 20 \text{ s} + 5 \text{ s} Ttotal=35 sT_{\text{total}} = 35 \text{ s}

  2. Calculate the average speed.

    Average speed=Total distanceTotal time\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} Average speed=550 m35 s\text{Average speed} = \frac{550 \text{ m}}{35 \text{ s}} Average speed≈15.71 m/s\text{Average speed} \approx \boxed{\text{15.71 m/s}}

    : Stone Dropped from a Cliff

    Given:
    Initial velocity (uu) = 0 m/s0 \text{ m/s} (since it's dropped)
    Height of the cliff (ss) = 80 m80 \text{ m}
    Acceleration due to gravity (gg) = 9.8 m/s29.8 \text{ m/s}^2

  3. Use the kinematic equation for displacement.

    s=ut+12gt2s = ut + \frac{1}{2}gt^2 80 m=(0 m/s)t+12(9.8 m/s2)t280 \text{ m} = (0 \text{ m/s})t + \frac{1}{2}(9.8 \text{ m/s}^2)t^2 80 m=4.9 m/s2⋅t280 \text{ m} = 4.9 \text{ m/s}^2 \cdot t^2

  4. Solve for t2t^2.

    t2=80 m4.9 m/s2t^2 = \frac{80 \text{ m}}{4.9 \text{ m/s}^2} t2≈16.3265 s2t^2 \approx 16.3265 \text{ s}^2

  5. Calculate tt.

    t=16.3265 s2t = \sqrt{16.3265 \text{ s}^2} t≈4.04 st \approx \boxed{\text{4.04 s}}
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