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: Understanding the Question:
We are given a position function $S(t)$ with constants $a$ and $b$. Using the kinematic boundary conditions (comes to rest at $t = 4\ \text{sec}$ when $S = 16\ \text{m}$), we need to find the constant acceleration of the particle.

Step 2: Key Formula or Approach:
Velocity ($v$) is the first derivative of distance with respect to time, and acceleration ($A$) is the second derivative: $$ v = \frac{\text{d}S}{\text{d}t}, \quad A = \frac{\text{d}v}{\text{d}t} $$

Step 3: Detailed Explanation:
Given displacement equation: $$ S = a t^2 + b t + 6 $$ Differentiating once to get the velocity function: $$ v = \frac{\text{d}S}{\text{d}t} = 2at + b $$ Differentiating a second time to determine acceleration: $$ A = \frac{\text{d}v}{\text{d}t} = 2a $$ Now let's apply our boundary conditions at $t = 4\ \text{s}$:

• The particle is at position $S = 16$: $$ 16 = a(4)^2 + b(4) + 6 \implies 16a + 4b = 10 \quad \text{--- (Equation 1)} $$

• The particle comes to rest ($v = 0$): $$ 0 = 2a(4) + b \implies b = -8a \quad \text{--- (Equation 2)} $$
Substitute Equation 2 into Equation 1: $$ 16a + 4(-8a) = 10 $$ $$ 16a - 32a = 10 \implies -16a = 10 \implies a = -\frac{10}{16} = -\frac{5}{8} $$ Now substitute the value of $a$ back into our acceleration formula: $$ A = 2a = 2 \times \left(-\frac{5}{8}\right) = -\frac{5}{4}\ \text{m/sec}^2 $$

Step 4: Final Answer:
The acceleration of the particle is $- \frac{5}{4}\ \text{m/sec}^2$, which matches option (D).