Question

Race car is accelerating and has a velocity of 10 m/s @ t=0. It completes a lap on a circular track of 400 m in 14 seconds. Calculate the components and magnitude of acceleration after the car has travelled 200 m. Assume acceleration along track is constant.

Answer 1

component of acceleration are a = 3.37 m/s² and ar = 22.74 m/s²

magnitude of acceleration is 22.98 m/s²

Step-by-step explanation:

given data

velocity = 10 m/s

initial time to = 0

distance s = 400 m

time t = 14 s

to find out

components and magnitude of acceleration after the car has travelled 200 m

solution

first we find the radius of circular track that is

we know distance S = 2πR

400 = 2πR

R = 63.66 m

and tangential acceleration is

S = ut + 0.5 ×at²

here u is initial speed and t is time and S is distance

400 = 10 × 14 + 0.5 ×a (14)²

a = 3.37 m/s²

and here tangential acceleration is constant

so velocity at distance 200 m

v² - u² = 2 a S

v² = 10² + 2 ( 3.37) 200

v = 38.05 m/s

so radial acceleration at distance 200 m

ar =
`(v^2)/(R)`

ar =
`(38.05^2)/(63.66)`

ar = 22.74 m/s²

so magnitude of total acceleration is

A =
`√(a^2 + ar^2)`

A =
`√(3.37^2 + 22.74^2)`

A = 22.98 m/s²

so magnitude of acceleration is 22.98 m/s²