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A motorcycle is moving at a constant velocity of 15 meters/second. Then it starts to accelerate and reaches a velocity of 24 meters/second in 3 seconds. What is the acceleration of the motorcycle over this time? Use $a=\frac{v-u}{t}$.


A$9 m / s ^2$
B$8 m / s ^2$
C$6 m / s ^2$
D$5 m / s ^2$
E$3 m / s ^2$

Verified Answer :

  • Identify initial velocity \(u = 15\text{ m/s}\), final velocity \(v = 24\text{ m/s}\), and time \(t = 3\text{ s}\).
  • Apply the constant acceleration formula \(a = \frac{v - u}{t}\).
  • Calculate the numerator: \(24 - 15 = 9\text{ m/s}\).
  • Divide by the time: \(\frac{9}{3} = 3\text{ m/s}^2\).
  • The correct acceleration is \(\boxed{3\text{ m/s}^2}\).

Explanation

  1. Analysis of Motion Parameters

We are given that a motorcycle starts at an initial velocity \(u = 15\text{ m/s}\) and accelerates to a final velocity \(v = 24\text{ m/s}\) over a duration of \(t = 3\text{ seconds}\). To find the acceleration \(a\), we use the kinematic formula for constant acceleration:
\[a = \frac{v - u}{t}\]
This formula calculates the rate of change of velocity per unit time.

  1. Calculation of Acceleration

To find the acceleration, we substitute the given values into the formula:
\[a = \frac{24\text{ m/s} - 15\text{ m/s}}{3\text{ s}}\]
First, we calculate the change in velocity in the numerator:
\[24 - 15 = 9\text{ m/s}\]
Next, we divide this change by the time interval:
\[a = \frac{9\text{ m/s}}{3\text{ s}} = 3\text{ m/s}^2\]
Thus, the acceleration of the motorcycle is \(3\text{ m/s}^2\).

  1. Conclusion and Final Answer

Comparing our result of \(3\text{ m/s}^2\) with the provided options:

  • \(9\text{ m/s}^2\)
  • \(8\text{ m/s}^2\)
  • \(6\text{ m/s}^2\)
  • \(5\text{ m/s}^2\)
  • \(3\text{ m/s}^2\)

The correct option is \(3\text{ m/s}^2\). Therefore, the final answer is \(\boxed{3\text{ m/s}^2}\).

Examples

Understanding acceleration is essential in everyday life, for instance, when driving a car. If you know how quickly your vehicle can speed up from a stop or increase speed to merge onto a highway, you can better judge distances and safety. This exact physical principle applies to everything from how a sprinter starts a race to how engineers design efficient transportation systems.

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