Question

A car starts from rest and moves around a circular track of radius 27.0 m. Its speed increases at the constant rate of 0.420 m/s2. (a) What is the magnitude of its net linear acceleration 13.0 s later

Answer 1

Answer:

1.18 m/s²

Explanation:

The linear speed = 13.0s

$v= v_i + at\\v = 0 + 0.42(13)\\v = 5.46m/s$

where v(i) is the initial linear speed

the radial acceleration of the car afte (t) = 13s is

$a_r = \frac{v^2}{r} \\= \frac{5.46^{2} }{27} \\= 1.104m/s^2$

the magnitude of the net acceleration is

$a_net = \sqrt{1.104^2 + 0.1764^2} \\= 1.18m/s^2$