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

NTA AbhyasNTA Abhyas 2020Laws of Motion

Solution:

Given $v=20 \, \text{m s}^{- 1}\text{,}$
$R = 100 \, \text{m}$
$\frac{d v}{d t}= \, \text{3} \, \text{m s}^{- 2}$
$a_{c}=\frac{v^{2}}{R}=\frac{\left(\right. 20 \left.\right)^{2}}{100}=4 \, \left(\text{m s}\right)^{- 2}$
$a_{t}=\frac{d v}{d t}=3 \, \text{m s}^{- 2}$
$a=\sqrt{a_{c}^{2} + a_{t}^{2}}=\sqrt{4^{2} + 3^{2}}=5 \, \text{m s}^{- 2}$