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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Ответ:
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
The maximum velocity is 39.06 m/s (nearest hundredth)
The graph shown below displays the velocity.
Ответ: