A car is moving on a circular path of radius 600 m such that the magnitudes of the tangential acceleration and centripetal acceleration are equal. The time taken by the car to complete first quarter of revolution, if it is moving with an initial speed of 54 km/hr is t(1-e^(-π/2))s. The value of t is ____________.

Solution

Physics sir · 5 steps

Turn the time derivative into a distance derivative, and the speed grows exponentially with arc length.

1. Set up the condition. The tangential acceleration is dv/dt and the centripetal acceleration is v²/R. Setting them equal gives dv/dt = v²/R, with the car speeding up.
2. Switch to distance. Write dv/dt = vdv/ds, where s is the arc length. Then dv/ds = v/R, so v = v₀ e^(s/R).
3. Find the time. Since dt = ds/v = e^(-s/R)/v₀ ds, integrate from s = 0 to a quarter turn, s = (π R)/2.

   t_(1/4) = ∫₀^(π R/2) e^(-s/R)/v₀ ds = R/v₀(1 - e^(-π/2))

4. Put in numbers. v₀ = 54 km/h = 15 m s⁻¹ and R = 600 m, so R/v₀ = 600/15 = 40 s.
5. So. Matching with t(1 - e^(-π/2)) gives t = 40.

The common slip: 11.1. Using 54 directly without converting km/h to m/s gives 600/54 ≈ 11.1. Convert to 15 m s⁻¹ first.

The idea to keep. When an acceleration depends on speed and the question asks about a distance or an angle, use dv/dt = vdv/ds to solve for v(s) first. Then integrate dt = ds/v.

Answer from the NTA final answer key. Solution by PrepWiser.