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Identify the initial velocity (u), final velocity (v), and time (t) for the first 4 seconds.

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3 parts
a

5 m/s2\text{5 m/s}^2

b

210 m\text{210 m}

c

8 m/s\text{8 m/s}

Step-by-step solution

8 steps

Here's the solution to the problem based on the provided speed-time graph:

Part a· 2 steps

Acceleration in the first 4 seconds.

  1. Identify the initial velocity (uu), final velocity (vv), and time (tt) for the first 4 seconds.

    The object starts from rest, so u=0 m/su = 0 \text{ m/s}.
    After 4 seconds, the speed is v=20 m/sv = 20 \text{ m/s}.
    The time taken is t=4 st = 4 \text{ s}.

  2. Calculate the acceleration using the formula a=v−uta = \frac{v - u}{t}.

    a=20 m/s−0 m/s4 sa = \frac{20 \text{ m/s} - 0 \text{ m/s}}{4 \text{ s}} a=20 m/s4 sa = \frac{20 \text{ m/s}}{4 \text{ s}} a=5 m/s2a = 5 \text{ m/s}^2

    The acceleration in the first 4 seconds is 5 m/s2\boxed{\text{5 m/s}^2}.

Part b

210 m\text{210 m}

Total distance covered by the object.

The total distance covered is the area under the speed-time graph. We can divide the graph into three sections: a triangle (0-4 s), a rectangle (4-10 s), and another triangle (10-15 s).

  1. Calculate the distance covered in the first 4 seconds (triangle).

    S1=12×base×heightS_1 = \frac{1}{2} \times \text{base} \times \text{height} S1=12×4 s×20 m/sS_1 = \frac{1}{2} \times 4 \text{ s} \times 20 \text{ m/s} S1=40 mS_1 = 40 \text{ m}

  2. Calculate the distance covered from 4 s to 10 s (rectangle).

    The duration is 10 s−4 s=6 s10 \text{ s} - 4 \text{ s} = 6 \text{ s}. The speed is constant at 20 m/s20 \text{ m/s}.

    S2=length×widthS_2 = \text{length} \times \text{width} S2=6 s×20 m/sS_2 = 6 \text{ s} \times 20 \text{ m/s} S2=120 mS_2 = 120 \text{ m}

  3. Calculate the distance covered from 10 s to 15 s (triangle).

    The duration is 15 s−10 s=5 s15 \text{ s} - 10 \text{ s} = 5 \text{ s}. The initial speed is 20 m/s20 \text{ m/s} and the final speed is 0 m/s0 \text{ m/s}.

    S3=12×base×heightS_3 = \frac{1}{2} \times \text{base} \times \text{height} S3=12×5 s×20 m/sS_3 = \frac{1}{2} \times 5 \text{ s} \times 20 \text{ m/s} S3=50 mS_3 = 50 \text{ m}

  4. Calculate the total distance.

    Stotal=S1+S2+S3S_{\text{total}} = S_1 + S_2 + S_3 Stotal=40 m+120 m+50 mS_{\text{total}} = 40 \text{ m} + 120 \text{ m} + 50 \text{ m} Stotal=210 mS_{\text{total}} = 210 \text{ m}

    The total distance covered by the object is 210 m\boxed{\text{210 m}}.

Part c

8 m/s\text{8 m/s}

Average speed in the last 4 seconds.

The last 4 seconds of the motion are from t=15 s−4 s=11 st = 15 \text{ s} - 4 \text{ s} = 11 \text{ s} to t=15 st = 15 \text{ s}.

  1. Determine the speed at t=11 st = 11 \text{ s} and t=15 st = 15 \text{ s}.

    At t=15 st = 15 \text{ s}, the object comes to rest, so v15=0 m/sv_{15} = 0 \text{ m/s}.
    To find the speed at t=11 st = 11 \text{ s}, we first calculate the deceleration in the last segment (10 s to 15 s).

    adecel=final speed−initial speedtime intervala_{\text{decel}} = \frac{\text{final speed} - \text{initial speed}}{\text{time interval}} adecel=0 m/s−20 m/s15 s−10 sa_{\text{decel}} = \frac{0 \text{ m/s} - 20 \text{ m/s}}{15 \text{ s} - 10 \text{ s}} adecel=−20 m/s5 s=−4 m/s2a_{\text{decel}} = \frac{-20 \text{ m/s}}{5 \text{ s}} = -4 \text{ m/s}^2

    Now, use v=u+atv = u + at for the interval from t=10 st=10 \text{ s} to t=11 st=11 \text{ s}.
    Here, u=20 m/su = 20 \text{ m/s} (speed at t=10 st=10 \text{ s}), a=−4 m/s2a = -4 \text{ m/s}^2, and t=11 s−10 s=1 st = 11 \text{ s} - 10 \text{ s} = 1 \text{ s}.

    v11=20 m/s+(−4 m/s2)(1 s)v_{11} = 20 \text{ m/s} + (-4 \text{ m/s}^2)(1 \text{ s}) v11=20 m/s−4 m/s=16 m/sv_{11} = 20 \text{ m/s} - 4 \text{ m/s} = 16 \text{ m/s}

  2. Calculate the average speed in the last 4 seconds.

    For uniformly changing speed, the average speed is initial speed+final speed2\frac{\text{initial speed} + \text{final speed}}{2}.

    vavg=v11+v152v_{\text{avg}} = \frac{v_{11} + v_{15}}{2} vavg=16 m/s+0 m/s2v_{\text{avg}} = \frac{16 \text{ m/s} + 0 \text{ m/s}}{2} vavg=16 m/s2=8 m/sv_{\text{avg}} = \frac{16 \text{ m/s}}{2} = 8 \text{ m/s}

    The average speed in the last 4 seconds is 8 m/s\boxed{\text{8 m/s}}.

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