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26 July, 13:29

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?

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  1. 26 July, 16:12
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    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°
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