Q6a. A Car starts from rest and accelerates uniformly for 20 second to attain a speed 25 ms^(-1). It maintains this speed for 30 second before decelerating uniformly to rest. The total time for the journey is 60second. (i) Sketch a velocity time graph for the motion (ii) use the graph to determine the (a) total distance Covered by the body (b) Deceleration of the Car

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Answers

2 parts
2

1125 m1125\ \text{m}

3

2.5 m/s22.5\ \text{m/s}^2

Step-by-step solution

9 steps

Part 1

(i) Sketch a velocity time graph for the motion

The velocity-time graph consists of three segments:

  1. Acceleration phase: A straight line from (0 s, 0 m/s) to (20 s, 25 m/s).
  2. Constant velocity phase: A horizontal line from (20 s, 25 m/s) to (50 s, 25 m/s).
  3. Deceleration phase: A straight line from (50 s, 25 m/s) to (60 s, 0 m/s).
Velocity (m/s)
^
|
25 +-------------------*
|                   / \
|                  /   \
|                 /     \
|                /       \
|               /         \
|              /           \
|             /             \
|            /               \
|           /                 \
0 +--------*-------------------*--------> Time (s)
  0       20                  50       60

Part 2

1125 m1125\ \text{m}

(ii) use the graph to determine the (a) total distance Covered by the body

  1. Identify the shape of the graph

    The velocity-time graph forms a trapezium. The total distance covered is the area under this graph.

  2. Calculate the area of the first triangle (acceleration phase)

    The first segment is a triangle from t=0s to t=20s, with a maximum velocity of 25 m/s.

    Area1=12×base×height=12×20 s×25 m/s=250 m\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20\ \text{s} \times 25\ \text{m/s} = 250\ \text{m}

  3. Calculate the area of the rectangle (constant velocity phase)

    The second segment is a rectangle from t=20s to t=50s, with a constant velocity of 25 m/s.

    Area2=length×width=(50 s−20 s)×25 m/s=30 s×25 m/s=750 m\text{Area}_2 = \text{length} \times \text{width} = (50\ \text{s} - 20\ \text{s}) \times 25\ \text{m/s} = 30\ \text{s} \times 25\ \text{m/s} = 750\ \text{m}

  4. Calculate the area of the second triangle (deceleration phase)

    The third segment is a triangle from t=50s to t=60s, with an initial velocity of 25 m/s and a final velocity of 0 m/s.

    Area3=12×base×height=12×(60 s−50 s)×25 m/s=12×10 s×25 m/s=125 m\text{Area}_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (60\ \text{s} - 50\ \text{s}) \times 25\ \text{m/s} = \frac{1}{2} \times 10\ \text{s} \times 25\ \text{m/s} = 125\ \text{m}

  5. Calculate the total distance

    Total Distance=Area1+Area2+Area3=250 m+750 m+125 m=1125 m\text{Total Distance} = \text{Area}_1 + \text{Area}_2 + \text{Area}_3 = 250\ \text{m} + 750\ \text{m} + 125\ \text{m} = 1125\ \text{m}

    The total distance covered by the car is 1125 m\boxed{1125\ \text{m}}.

Part 3

2.5 m/s22.5\ \text{m/s}^2

(ii) use the graph to determine the (b) Deceleration of the Car

  1. Identify the deceleration phase on the graph

    The deceleration phase occurs from t=50s to t=60s, where the velocity decreases from 25 m/s to 0 m/s.

  2. Determine the change in velocity and time interval

    Initial velocity (uu) at t=50s = 25 m/s
    Final velocity (vv) at t=60s = 0 m/s
    Time interval (Δt\Delta t) = 60 s - 50 s = 10 s

  3. Calculate the acceleration

    Acceleration is the gradient (slope) of the velocity-time graph.

    a=ΔvΔt=v−uΔta = \frac{\Delta v}{\Delta t} = \frac{v - u}{\Delta t} a=0 m/s−25 m/s10 s=−25 m/s10 s=−2.5 m/s2a = \frac{0\ \text{m/s} - 25\ \text{m/s}}{10\ \text{s}} = \frac{-25\ \text{m/s}}{10\ \text{s}} = -2.5\ \text{m/s}^2

  4. State the deceleration

    Deceleration is the magnitude of the negative acceleration.

    Deceleration=∣a∣=∣−2.5 m/s2∣=2.5 m/s2\text{Deceleration} = |a| = |-2.5\ \text{m/s}^2| = 2.5\ \text{m/s}^2

    The deceleration of the car is 2.5 m/s2\boxed{2.5\ \text{m/s}^2}.

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