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

1.18 m/s²

Step-by-step 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 = (v^2)/(r) \\= (5.46^(2) )/(27) \\= 1.104m/s^2`

the magnitude of the net acceleration is

`a_net = √(1.104^2 + 0.1764^2) \\= 1.18m/s^2`