A motor car is moving with speed 30 m s -1 on a circular path of radius 500 m. Its speed is increasing at the rate of 2 m s -2, what will be its resultant acceleration?

Question

A motor car is moving with speed 30 m s -1 on a circular path of radius 500 m. Its speed is increasing at the rate of 2 m s -2, what will be its resultant acceleration? (A) 2.5 m s -2 (B) 2.7 m s -2 (C) 2 m s -2 (D) 4.5 m s -2

Tardigrade Admin · Accepted Answer

Centripetal acceleration, ac=(v2/r)=((30)2/500)=(9/5)=1.8 m s -2 Tangential acceleration, at=2 m s -2 ∴ Resultant acceleration, a=√ac2+at2=√(1.8)2+22=2.7 m s -2

Q. A motor car is moving with speed $30 m s^{- 1}$ on a circular path of radius $500 m$ . Its speed is increasing at the rate of $2 m s^{- 2},$ what will be its resultant acceleration?

$2.5 m s^{- 2}$

$2.7 m s^{- 2}$

$2 m s^{- 2}$

$4.5 m s^{- 2}$

Centripetal acceleration,
$a_{c} = \frac{v ^{2}}{r} = \frac{( 30 ) ^{2}}{500} = \frac{9}{5} = 1.8 m s^{- 2}$
Tangential acceleration, $a_{t} = 2 m s^{- 2}$
$∴$ Resultant acceleration,
$a = a_{c}^{2} + a_{t}^{2} = (1.8)^{2} + 2^{2} = 2.7 m s^{- 2}$

Q. A motor car is moving with speed 30ms−1 on a circular path of radius 500m. Its speed is increasing at the rate of 2ms−2, what will be its resultant acceleration?

2.5ms−2

2.7ms−2

2ms−2

4.5ms−2