Solveeit Logo

Question

Physics Question on laws of motion

A car is moving with speed 20m s120 \, \, \text{m s}^{- 1} on a circular path of radius 100m100 \, \, \text{m} . Its speed is increasing at a rate of 3m s23 \, \text{m s}^{- 2} . The magnitude of the acceleration of the car at that moment is

A

1m s21\text{m s}^{- 2}

B

3m s23\text{m s}^{- 2}

C

4m s24\text{m s}^{- 2}

D

5m s25\text{m s}^{- 2}

Answer

5m s25\text{m s}^{- 2}

Explanation

Solution

Given v=20m s1,v=20 \, \text{m s}^{- 1}\text{,} R=100mR = 100 \, \text{m} dvdt=3m s2\frac{d v}{d t}= \, \text{3} \, \text{m s}^{- 2} ac=v2R=(20)2100=4(m s)2a_{c}=\frac{v^{2}}{R}=\frac{\left(\right. 20 \left.\right)^{2}}{100}=4 \, \left(\text{m s}\right)^{- 2} at=dvdt=3m s2a_{t}=\frac{d v}{d t}=3 \, \text{m s}^{- 2} a=ac2+at2=42+32=5m s2a=\sqrt{a_{c}^{2} + a_{t}^{2}}=\sqrt{4^{2} + 3^{2}}=5 \, \text{m s}^{- 2}