Question

The acceleration of a motorcycle is given by ax(t)=At-Bt2,A=1.5m/s2,B=0.120m/s2,the motorcycle at rest at t=0 a)Find its position and velocity as function of time b)What maximum velocity it attains

Answer 1

(a)

`v(t) = (At^2)/(2) -(Bt^3)/(3)` --- velocity

`x(t) = (At^3)/(6) -(Bt^4)/(12) + x(0)` --- position

(b) The maximum velocity is: 39.0625m/s

Step-by-step explanation:

Given

`a_x(t) =At -Bt^2`

`A = 1.5m/s^3`

`B = 0.120m/s^4`

Solving (a): The position and velocity as a function

To calculate the change in velocity, we integrate the given acceleration

i.e.

`\int\limits^(v(t))_(v(0)) {dv} = \int\limits^t_0 {a(t)} \, dt`

Integrate

`v|\limits^(v(t))_(v(0)) = \int\limits^t_0 {a(t)} \, dt`

Substitute v(t) and v(0) for v

`v(t) - v(0) = \int\limits^t_0 {a(t)} \, dt`

`v(0) = 0` ---- The initial velocity.

So, we have:

`v(t) - 0 = \int\limits^t_0 {a(t)} \, dt`

`v(t) = \int\limits^t_0 {a(t)} \, dt`

Substitute:
`a_x(t) =At -Bt^2`

`v(t) = \int\limits^t_0 {[At -Bt^2]} \, dt`

Integrate

`v(t) = {(At^2)/(2) -(Bt^3)/(3)}|\limits^t_0`

Substitute 0 and t for t

`v(t) = [(At^2)/(2) -(Bt^3)/(3)] - [(A*0^2)/(2) -(B*0^3)/(3)]`

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

`v(t) = (At^2)/(2) -(Bt^3)/(3)`

To calculate the change in position, we integrate the calculated velocity

i.e.

`\int\limits^(x(t))_(x(0)) {dx} = \int\limits^t_0 {v(t)} \, dt`

Integrate

`x(t)|\limits^(x(t))_(x(0)) = \int\limits^t_0 {v(t)} \, dt`

Substitute x(t) and x(0) for x(t)

`x(t) - x(0) = \int\limits^t_0 {v(t)} \, dt`

Substitute:
`v(t) = (At^2)/(2) -(Bt^3)/(3)`

`x(t) - x(0) = \int\limits^t_0 {(At^2)/(2) -(Bt^3)/(3)} \, dt`

Integrate

`x(t) - x(0) = (At^3)/(6) -(Bt^4)/(12)}|\limits^t_0`

Substitute t and 0 for t

`x(t) - x(0) = [(At^3)/(6) -(Bt^4)/(12)] - [(A*0^3)/(6) -(B*0^4)/(12)]`

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

`x(t) - x(0) = (At^3)/(6) -(Bt^4)/(12)`

Make x(t) the subject

`x(t) = (At^3)/(6) -(Bt^4)/(12) + x(0)`

Where x(0) represents the initial position

Solving (b): The maximum velocity

First, we calculate the time at which it attains the maximum height

Set acceleration to 0

`a_x(t) =At -Bt^2`

`At - Bt^2 = 0`

Factorize

`t(A - Bt) = 0`

Split

`t = 0\ or\ A - Bt = 0`

t can't be 0.

So, we have:

`A - Bt = 0`

Solve for t

`-Bt = -A`

`t = (-A)/(-B)`

`t = (A)/(B)`

Substitute:
`A = 1.5m/s^3` and
`B = 0.120m/s^4`

`t = (1.5m/s^3)/(0.120m/s^4)`

`t = (1.5)/(0.120)s`

`t = 12.5s`

Substitute
`t = 12.5s` in v(t) to get the maximum velocity

`v(t) = (At^2)/(2) -(Bt^3)/(3)`

`v(t) = (A*12.5^2)/(2) -(B*12.5^3)/(3)`

`v(t) = (156.25A)/(2) -(1953.125B)/(3)`

Substitute:
`A = 1.5m/s^3` and
`B = 0.120m/s^4`

`vmax = (156.25*1.5)/2 -(1953.125*0.120)/3`

`v_(max) = 117.1875 - 78.125`

`v_(max) = 39.0625`