Question

A car is travelling at 40 m/s on a circular path of radius 40 m. It is increasing its speed at the rate of \(2 \text{m/s}^2\). Its net acceleration is (in \(\text{m/s}^2\)) nearly

• A. \(4 \text{m/s}^2\)
• B. \(8 \text{m/s}^2\)
• C. \(16 \text{m/s}^2\)
• D. \(40 \text{m/s}^2\)

Hint:

Combine the centripetal and tangential accelerations at right angles.

Answer 1

The Correct Option is D

Step 1: Approach
Compare sizes first to see which term dominates.

Step 2: Sizes
$a_c=v^2/r=1600/40=40$ and $a_t=2$. The ratio $a_t/a_c=0.05$, so the tangential part is small.

Step 3: Combine
$a=a_c\sqrt{1+(0.05)^2}=40\times1.00125=40.05$.

Step 4: Answer
This is nearly 40 m/s$^2$, option (D). The smaller options 4, 8 and 16 would be below the centripetal part alone, which cannot be.

Final Answer:
The centripetal part is 40 and the tangential part is 2, so the net is about 40 m/s squared, option (D). \[ \boxed{\approx 40\ \text{m/s}^2} \]