Question

A car starts from rest and moves around a circular track of radius 26.0 m. Its speed increases at the constant rate of 0.570 m/s2. (a) What is the magnitude of its net linear acceleration 12.0 s later? (b) What angle does this net acceleration vector make with the car's velocity at this time?

Answer 1

Answer:

a) 1.49m/s²

b) 67.5°

Explanation:

Given

Initial angular speed of the car, ω = 0 rad/s

Radius of the circular track, r = 26m

Linear tangential acceleration is the car, a(t) = 0.57m/s²

Linear speed of the car after 12s =

V = u + at

V = 0 + 0.5*12

V = 6m/s

Radial acceleration of the car after t = 12s is

a(r) = v²/r

a(r) = 6²/26

a(r) = 1.38m/s²

Magnitude of the net acceleration of the car after t = 12s is

a(net) = √[a(r)² + a(t)²]

a(net) = √[1.38² + 0.57²]

a(net) = √2.2293

a(net) = 1.49m/s²

It should be noted that, the direction of the cars magnitude is always the same with the direction of the cars tangential acceleration.

the angle Φ between the tangential and net acceleration is

Φ = cos^-1 a(t)/a(net)

Φ = cos^-1 0.57/1.49

Φ = cos^-1 0.3826

Φ = 67.5°