A racecar, moving at a constant tangential speed of 60 m/s, takes one lap around a circular track in 50 seconds. Determine the magnitude of the acceleration of the car.

Answer 1

The tangential speed of the car is given as 60 m/s, which means that the car's velocity is constant along the tangent to the circular track.

The time taken for one lap around the track is 50 seconds, so the period (T) of the car's motion is 50 seconds.

The formula for acceleration in circular motion is given by:
a = v^2 / r

where:
a = acceleration
v = tangential speed
r = radius of the circular track

Given that v = 60 m/s and T = 50 s, we first need to find the radius of the circular track using the formula:
v = 2πr / T
60 = 2πr / 50
r = 60*50 / (2*π) = 150.96 m

Now we can calculate the acceleration:
a = v^2 / r
a = 60^2 / 150.96
a = 3600 / 150.96 ≈ 23.87 m/s^2

Therefore, the magnitude of the acceleration of the racecar is approximately 23.87 m/s^2.