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A) Find the acceleration of the object when t=3. b) Given that the object decelerates at 2 m/s2, find the value of t.

Physics

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: Moving object

Part a· 2 steps

Find the acceleration of the object when t=3t=3.

  1. Identify the segment of the graph.

    From t=0t=0 to t=5t=5 seconds, the speed-time graph is a straight line. The acceleration is constant in this interval. The points on the graph for this segment are (0,0)(0,0) and (5,24)(5,24).

  2. Calculate the acceleration.

    Acceleration is the gradient of the speed-time graph.

    Acceleration=Change in speedChange in time=24 m/s−0 m/s5 s−0 s\text{Acceleration} = \frac{\text{Change in speed}}{\text{Change in time}} = \frac{24\, \text{m/s} - 0\, \text{m/s}}{5\, \text{s} - 0\, \text{s}} Acceleration=245 m/s2=4.8 m/s2\text{Acceleration} = \frac{24}{5}\, \text{m/s}^2 = 4.8\, \text{m/s}^2

    The acceleration when t=3t=3 is 4.8 m/s2\boxed{\text{4.8 m/s}^2}.

Part b· 3 steps

Given that the object decelerates at 2 m/s22\, \text{m/s}^2, find the value of tt.

  1. Identify the deceleration segment.

    Deceleration occurs in the last segment of the graph, from t=15t=15 seconds to tt. At t=15t=15 seconds, the speed is 24 m/s24\, \text{m/s}. At time tt, the speed is 0 m/s0\, \text{m/s} (as the graph ends on the time axis).
    Deceleration of 2 m/s22\, \text{m/s}^2 means the acceleration is −2 m/s2-2\, \text{m/s}^2.

  2. Use the acceleration formula to find tt.

    Acceleration=Final speed−Initial speedFinal time−Initial time\text{Acceleration} = \frac{\text{Final speed} - \text{Initial speed}}{\text{Final time} - \text{Initial time}} −2 m/s2=0 m/s−24 m/st s−15 s-2\, \text{m/s}^2 = \frac{0\, \text{m/s} - 24\, \text{m/s}}{t\, \text{s} - 15\, \text{s}} −2=−24t−15-2 = \frac{-24}{t - 15}

  3. Solve for tt.

    −2(t−15)=−24-2(t - 15) = -24 t−15=−24−2t - 15 = \frac{-24}{-2} t−15=12t - 15 = 12 t=12+15t = 12 + 15 t=27 st = 27\, \text{s}

    The value of tt is 27 s\boxed{\text{27 s}}.

Part c· 2 steps

Calculate the average speed of the object in the first 10 seconds of the journey.

  1. Calculate the total distance covered in the first 10 seconds.

    The distance is the area under the speed-time graph.
    From t=0t=0 to t=5t=5 seconds, the shape is a triangle:

    Area1=12×base×height=12×5 s×24 m/s=60 m\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5\, \text{s} \times 24\, \text{m/s} = 60\, \text{m}

    From t=5t=5 to t=10t=10 seconds, the shape is a rectangle:

    Area2=width×height=(10 s−5 s)×24 m/s=5 s×24 m/s=120 m\text{Area}_2 = \text{width} \times \text{height} = (10\, \text{s} - 5\, \text{s}) \times 24\, \text{m/s} = 5\, \text{s} \times 24\, \text{m/s} = 120\, \text{m}

    Total distance = Area1+Area2=60 m+120 m=180 m\text{Area}_1 + \text{Area}_2 = 60\, \text{m} + 120\, \text{m} = 180\, \text{m}.

  2. Calculate the average speed.

    Average speed is total distance divided by total time.

    Average speed=Total distanceTotal time=180 m10 s=18 m/s\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{180\, \text{m}}{10\, \text{s}} = 18\, \text{m/s}

    The average speed in the first 10 seconds is $\boxed{\text{*18 m/

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