Problem

A motorcycle starts from rest and accelerates  as shown in the figure. Determine (a) the m...

A motorcycle starts from rest and accelerates  as shown in the figure. Determine (a) the motorcycle’s speed at t = 4.00 s and at t = 14.0 s, and (b)  the distance  traveled in the first 14.0 s.

Step-by-Step Solution

Solution 1

This problem can be solved by making use of the values shown in the graph below.

The graph between acceleration and time:

C:\Users\admin\Documents\My Data Sources\13326-2-53E.png

(a)

The speed of the motor cycle at \(t=4.00 \mathrm{~s}\) is calculated by using the values in the above graph. The initial velocity of the motor cycle, \(v_{0}\) is \(0 \mathrm{~m} / \mathrm{s}\), time interval \(t\) is \(4.00 \mathrm{~s}\) and acceleration of the motorcycle \(a\) is \(5 \mathrm{~m} / \mathrm{s}^{2}\).

The speed of the motorcycle is,

$$ \begin{aligned} v_{\mathrm{att}=4.0 \mathrm{~s}} &=v_{0}+a t \\ &=0+\left(5 \mathrm{~m} / \mathrm{s}^{2}\right)(4.0 \mathrm{~s}) \\ &=20 \mathrm{~m} / \mathrm{s} \end{aligned} $$

Thus, the speed of the motorcycle is \(20 \mathrm{~m} / \mathrm{s}\).

The speed is constant between \(t=4.0 \mathrm{~s}\) and \(t=12.0 \mathrm{~s} . \mathrm{So}\), at \(t=12.00 \mathrm{~s}\), the speed of the motorcycle is \(v=20 \mathrm{~m} / \mathrm{s}\)

To determine the speed at \(t=14.00 \mathrm{~s}\), let us consider the time interval from \(t=12.0 \mathrm{~s}\) to \(t=14.0 \mathrm{~s}\), and the acceleration of the motorcycle is this time interval is, \(a=-4 \mathrm{~m} / \mathrm{s}^{2}\).

The speed of the motorcycle is,

$$ \begin{aligned} v_{\text {at t= } 14.0 s} &=v_{0}+a t \\ &=20 \mathrm{~m} / \mathrm{s}+\left(-4 \mathrm{~m} / \mathrm{s}^{2}\right)(14.0 \mathrm{~s}-12.0 \mathrm{~s}) \\ &=12 \mathrm{~m} / \mathrm{s} \end{aligned} $$

Thus, the speed of the motorcycle is \(12 \mathrm{~m} / \mathrm{s}\).

(b)

To determine the distance traveled in the first \(14.0 \mathrm{~s}\), Calculate the distance in the subsequent time intervals and then sum all the terms and the total distance is.

Distance traveled by the motorcycle from \(t=0 \mathrm{~s}\) to \(t=4.0 \mathrm{~s}\).

$$ \begin{aligned} s_{1} &=v_{0} t+\frac{1}{2} a t^{2} \\ &=0+\frac{1}{2}\left(5 \mathrm{~m} / \mathrm{s}^{2}\right)(4.0 \mathrm{~s})^{2} \\ &=40 \mathrm{~m} \end{aligned} $$

Distance traveled by the motorcycle from \(t=4 \mathrm{~s}\) to \(t=12.0 \mathrm{~s}\),

$$ \begin{aligned} s_{2} &=v t \\ &=(20 \mathrm{~m} / \mathrm{s})(12.0 \mathrm{~s}-4.0 \mathrm{~s}) \\ &=160 \mathrm{~m} \end{aligned} $$

Distance traveled by the motorcycle from \(t=12 \mathrm{~s}\) to \(t=14.0 \mathrm{~s}\),

$$ \begin{aligned} s_{3} &=v_{0} t+\frac{1}{2} a t^{2} \\ &=(20 \mathrm{~m} / \mathrm{s})(14.0 \mathrm{~s}-12.0 \mathrm{~s})+\frac{1}{2}\left(-4 \mathrm{~m} / \mathrm{s}^{2}\right)(14.0 \mathrm{~s}-12.0 \mathrm{~s})^{2} \\ &=32 \mathrm{~m} \end{aligned} $$

Distance traveled in the first \(14.0 \mathrm{~s}\) is

$$ \begin{aligned} s &=s_{1}+s_{2}+s_{3} \\ &=40 \mathrm{~m}+160 \mathrm{~m}+32 \mathrm{~m} \\ &=232 \mathrm{~m} \end{aligned} $$

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