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.

Respuesta :

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.
Ver imagen Аноним

The maximum velocity it attains is 39.1 m/s

Further explanation

The velocity is changing over the course of time. Velocity is the rate of motion in a specific direction. Whereas acceleration is the rate of change of velocity of an object with respect to time. Maximum velocity is reached when you stop accelerating, To calculate velocity using acceleration, we start by multiplying the acceleration by the change in time

The acceleration of a motorcycle:

where [tex]a = 1.50 \frac{m}{s^{3}}[/tex] and [tex]ax* (t) = at - b*t^{2}[/tex]

[tex]a = 0.120  \frac{m}{s^{4}}[/tex]

The motorcycle is at rest at the origin at time t=0.

The maximum velocity it attains = ?

Answer:

[tex]v(t) = (1/2)At^2 - (1/3)Bt^3 v(0)[/tex]

but v(0) = 0.

[tex]x(t) = (1/6)At^3 - (1/12)Bt^4 x(0)[/tex]

but x(0) = 0.

[tex]v(t) = (0.75 \frac{m}{s^{3}}) t^2 - (0.04 \frac{m}{s^{4}})t^3[/tex]

Next step is find its position as a function of time.

[tex]x(t)= x(t) = (0.25 \frac{m}{s^{3}})t^3 - (0.01 \frac{m}{s^{4}})t^4[/tex]

Then, calculate the maximum velocity it attains.

Max velocity will be attained when

[tex]At = Bt^2[/tex]

T= 1.50/0.120 = 12.5 seconds

V(t) = 0.750(12.5)^2 – 0.04(12.5)^3 = 39.1 m/s

Learn more

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Answer details

Grade:  9

Subject:  physics

Chapter:  the maximum velocity

Keywords: the maximum velocity