Question

A particle is moving on a straight line. The distance S travelled in time t is given by $S = a t^2 + b t + 6$. If the particle comes to rest after 4 seconds at a distance of 16 m from the starting point, then the acceleration of the particle is

• A. $- \frac{3}{4}\ \text{m/sec}^2$
• B. $- \frac{1}{2}\ \text{m/sec}^2$
• C. $- 1\ \text{m/sec}^2$
• D. $- \frac{5}{4}\ \text{m/sec}^2$

Hint:

Since acceleration is constant ($2a$), this is a uniform acceleration problem. You can quickly cross-verify using standard equations of motion: $S_{\text{net}} = v_{\text{avg}} \times t$. Since it ends at rest, the initial velocity must be twice the average, allowing you to solve for $a$ fast!

Answer 1

The Correct Option is D

Step 1: Differentiate.
From $S = at^2 + bt + 6$, velocity is $v = 2at + b$ and acceleration is $2a$, a constant.

Step 2: Apply the conditions.
At $t = 4$, at rest means $8a + b = 0$, so $b = -8a$. At $t = 4$, $S = 16$ gives $16a + 4b = 10$.

Step 3: Solve.
Substituting, $16a - 32a = 10$, so $a = -\dfrac{5}{8}$ and acceleration $= 2a = -\dfrac{5}{4}$ m/sec$^2$.
\[ \boxed{-\dfrac{5}{4}\ \text{m/sec}^2,\ \text{option 4}} \]