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

Step 1: Picture the motion.
The particle speeds up from rest, reaches a top speed $v_{\max}$, then slows down to rest. The same peak speed marks the end of speeding up and the start of slowing down.

Step 2: Name the two times.
Let $t_1$ be the time spent accelerating and $t_2$ the time spent decelerating, with \[ t_1+t_2=10\ \text{s}. \]

Step 3: Express the peak speed two ways.
Starting from rest, $v_{\max}=\alpha t_1=6t_1$. Ending at rest, $v_{\max}=\beta t_2=4t_2$.

Step 4: Relate $t_1$ and $t_2$.
From $6t_1=4t_2$ we get $t_2=\dfrac{3}{2}t_1$.

Step 5: Use the total time.
\[ t_1+\frac{3}{2}t_1=10\ \Rightarrow\ \frac{5}{2}t_1=10\ \Rightarrow\ t_1=4\ \text{s}. \]

Step 6: Find the peak speed.
\[ v_{\max}=6t_1=6\times4=24\ \text{m s}^{-1}. \]
\[ \boxed{v_{\max}=24\ \text{m s}^{-1}} \]