The velocity of a car travelling on a straight road is 36 kmh-1 at an instant of time. Now travelling with uniform acceleration for 10 s, the velocity becomes exactly double. If the wheel radius of the car is 25 cm, then which of the following numbers is the closest to the number of revolutions that the wheel makes during this 10 s?

Question

The velocity of a car travelling on a straight road is 36 kmh-1 at an instant of time. Now travelling with uniform acceleration for 10 s, the velocity becomes exactly double. If the wheel radius of the car is 25 cm, then which of the following numbers is the closest to the number of revolutions that the wheel makes during this 10 s? (A) 84 (B) 95 (C) 126 (D) 135

Tardigrade Admin · Accepted Answer

We have, θ=2 π n=(((vf2/r2)-(vi2/r2))/(2 (a)r)) . ⇒ n=(vf2-vi2/(2 a r)(2 π)) ≈ 95

84

95

126

135

We have, θ=2πn=(2ra​)(r2vf2​​−r2vi2​​)​. ⇒n=(2ar)(2π)vf2​−vi2​​≈95

We have, $θ = 2 π n = \frac{( \frac{v _{f}^{2}}{r ^{2}} - \frac{v _{i}^{2}}{r ^{2}} )}{( 2 \frac{a}{r} )} .$
$\Rightarrow n = \frac{v _{f}^{2} - v _{i}^{2}}{( 2 a r ) ( 2 π )} \approx 95$