Question

A particle accelerates from rest at constant rate \[ \alpha=6\ \text{m s}^{-2} \] for some time, after which it decelerates at constant rate \[ \beta=4\ \text{m s}^{-2} \] to come to rest. If the total time of travel is \(10\) seconds, what is the maximum speed attained by the particle during its motion?

• A. \(6\ \text{m s}^{-1}\)
• B. \(12\ \text{m s}^{-1}\)
• C. \(24\ \text{m s}^{-1}\)
• D. \(12.5\ \text{m s}^{-1}\)

Hint:

When a particle accelerates and then decelerates to rest, the peak speed is common to both phases: \[ a t_1=\beta t_2. \] Use this relation together with the total time condition.

Answer 1

The Correct Option is C

Concept: Let \[ t_1 \] be the time of acceleration and \[ t_2 \] be the time of deceleration. Then \[ t_1+t_2=10. \] The maximum speed attained is the same at the end of acceleration and at the beginning of deceleration.

Step 1: Express maximum speed in two ways. During acceleration, \[ v_{\max}=6t_1. \] During deceleration, \[ v_{\max}=4t_2. \] Hence, \[ 6t_1=4t_2. \] \[ 3t_1=2t_2. \] \[ t_2=\frac32 t_1. \]

Step 2: Use the total time. \[ t_1+t_2=10 \] \[ t_1+\frac32 t_1=10 \] \[ \frac52 t_1=10 \] \[ t_1=4\ \text{s}. \] Therefore, \[ t_2=6\ \text{s}. \]

Step 3: Calculate the maximum speed. \[ v_{\max}=6t_1 \] \[ =6\times4 \] \[ =24\ \text{m s}^{-1}. \] \[\begin{aligned} \boxed{24\ \text{m s}^{-1}} \end{aligned}\] Hence, option \(\mathbf{(C)}\) is correct.