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

Answer:

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

magnitude of acceleration is  22.98 m/s²

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 = $\frac{v^2}{R}$

ar = $\frac{38.05^2}{63.66}$

ar = 22.74 m/s²

so magnitude of total acceleration is

A = $\sqrt{a^2 + ar^2}$

A = $\sqrt{3.37^2 + 22.74^2}$

A = 22.98 m/s²

so magnitude of acceleration is  22.98 m/s²