The acceleration of a motorcycle is given by ax (t) = at-bt2, where a=1.50m/s3 and b=0.120m/s4. the motorcycle is at rest at the origin at time t=0. calculate the maximum velocity it attains.

Question

Physics

Belinda

6 August, 17:10

The acceleration of a motorcycle is given by ax (t) = at-bt2, where a=1.50m/s3 and b=0.120m/s4. the motorcycle is at rest at the origin at time t=0. calculate the maximum velocity it attains.

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Answers (1)

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Leon Mcintosh · Answer 1

From given information, the acceleration is

a (t) = 1.5t - 0.12t² m/s²

Integrate to obtain the velocity.

v (t) = (1/2) * 1.5t² - (1/3) * 0.12t³ + c₁

= 0.75t² - 0.04t³ + c₁ m/s

Because v (0) = 0 (given), therefore c₁ = 0

The velocity is

v (t) = 0.75t² - 0.04t³ m/

The velocity is maximum when the acceleration is zero. That is,

t (1.5 - 0.12t) = 0

t = 0 or t = 1.5/.12 = 12.5 s

Reject t = 0 because it yields zero value.

The maximum velocity is

v (12.5) = 0.75 * (12.5²) - 0.04 * (12.5³) = 39.0625 m/s

Answer: The maximum velocity is 39.06 m/s (nearest hundredth)

The graph shown below displays the velocity.