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Calculate the acceleration shown by each of the graphs below.

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3 parts
a

15 km/h2\text{15 km/h}^2

b

0 m/s2\text{0 m/s}^2

c

-6.25 m/s2\text{-6.25 m/s}^2

Step-by-step solution

9 steps

To calculate acceleration from a velocity-time graph, we find the gradient (slope) of the line. The formula for acceleration is:

a=ΔvΔt=final velocity−initial velocityfinal time−initial timea = \frac{\Delta v}{\Delta t} = \frac{\text{final velocity} - \text{initial velocity}}{\text{final time} - \text{initial time}}

Part a· 3 steps

  1. Identify initial and final points from the graph.

    The graph starts at (t1,v1)=(0 hours,0 km/h)(t_1, v_1) = (0 \text{ hours}, 0 \text{ km/h}) and ends at (t2,v2)=(2 hours,30 km/h)(t_2, v_2) = (2 \text{ hours}, 30 \text{ km/h}).

  2. Calculate the change in velocity and change in time.

    Δv=30 km/h−0 km/h=30 km/h\Delta v = 30 \text{ km/h} - 0 \text{ km/h} = 30 \text{ km/h} Δt=2 hours−0 hours=2 hours\Delta t = 2 \text{ hours} - 0 \text{ hours} = 2 \text{ hours}

  3. Calculate the acceleration.

    a=30 km/h2 hours=15 km/h2a = \frac{30 \text{ km/h}}{2 \text{ hours}} = 15 \text{ km/h}^2

    The acceleration for graph a) is 15 km/h2\boxed{\text{15 km/h}^2}

Part b

0 m/s2\text{0 m/s}^2

  1. Identify initial and final points from the graph.

    The graph shows a constant velocity. Let's take (t1,v1)=(0 s,15 m/s)(t_1, v_1) = (0 \text{ s}, 15 \text{ m/s}) and (t2,v2)=(4 s,15 m/s)(t_2, v_2) = (4 \text{ s}, 15 \text{ m/s}).

  2. Calculate the change in velocity and change in time.

    Δv=15 m/s−15 m/s=0 m/s\Delta v = 15 \text{ m/s} - 15 \text{ m/s} = 0 \text{ m/s} Δt=4 s−0 s=4 s\Delta t = 4 \text{ s} - 0 \text{ s} = 4 \text{ s}

  3. Calculate the acceleration.

    a=0 m/s4 s=0 m/s2a = \frac{0 \text{ m/s}}{4 \text{ s}} = 0 \text{ m/s}^2

    The acceleration for graph b) is 0 m/s2\boxed{\text{0 m/s}^2}

Part c

-6.25 m/s2\text{-6.25 m/s}^2

  1. Identify initial and final points from the graph.

    The graph starts at (t1,v1)=(0 s,25 m/s)(t_1, v_1) = (0 \text{ s}, 25 \text{ m/s}) and ends at (t2,v2)=(4 s,0 m/s)(t_2, v_2) = (4 \text{ s}, 0 \text{ m/s}).

  2. Calculate the change in velocity and change in time.

    Δv=0 m/s−25 m/s=−25 m/s\Delta v = 0 \text{ m/s} - 25 \text{ m/s} = -25 \text{ m/s} Δt=4 s−0 s=4 s\Delta t = 4 \text{ s} - 0 \text{ s} = 4 \text{ s}

  3. Calculate the acceleration.

    a=−25 m/s4 s=−6.25 m/s2a = \frac{-25 \text{ m/s}}{4 \text{ s}} = -6.25 \text{ m/s}^2

    The acceleration for graph c) is -6.25 m/s2\boxed{\text{-6.25 m/s}^2}

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