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: Understanding the Concept
A car moving on a circle while speeding up has two perpendicular accelerations: the centripetal one pointing to the centre and the tangential one along the motion.

Step 2: Centripetal acceleration
\[ a_c=\frac{v^2}{r}=\frac{40^2}{40}=40\ \text{m/s}^2 \]

Step 3: Tangential acceleration
\[ a_t=2\ \text{m/s}^2 \]

Step 4: Net acceleration
The two are at right angles, so
\[ a=\sqrt{a_c^2+a_t^2}=\sqrt{1600+4}=\sqrt{1604}\approx40.05\ \text{m/s}^2 \]

Step 5: Conclusion
The value is nearly 40 m/s\(^2\), option (D). The tangential part is tiny compared with the centripetal part, so it changes the answer only in the second decimal place.

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