Suppose that a car starts from rest with its engine providin

Suppose that a car starts from rest with its engine providing a constant acceleration of 25 ft/s^2. If air resistance provides (.0004)v^2 ft/s^2 of deceleration, find the following: A. The car\'s maximum velocity B. How long the car takes to attain 90% of its limiting velocity C. How far it travels in doing so

Solution

I have solved this question earlier with different figures. Please workout using yours figures. If you need any further help just PM me. If I have helped you please rate me 5 stars first (before you rate anyone else)

Suppose that a car starts from rest with its engine providing a constant acceleration of 13.5 ft/s^2. If air resistance provides (.0006)v^2 ft/s^2 of deceleration, find the following,

a) The cars maximum possible (limiting) velocity

b)How long the car takes to attain 90% of its limiting velocity

c) How far it travels in doing so.

Answer

Acceleration = x\'\' = v\' = 13.5 - 0.0006 v^2

v\' = dv/dt

So, dv / (13.5- 0.0006 v^2) = dt

(1/0.0006) *dv / (13.5/0.0006 - v^2) = dt

(1/0.0006) dv / [ (sqrt(13.5/0.0006))^2 - v^2] = dt

This can be compared with standard formula Integral dx/(a^2 - x^2) = -1/(2a) ln [(a-x)/(a+x)]

We have a = sqrt(13.5/0.0006) = 150

Thus, -(1/0.0006) (1/(2*150)) ln [(150 - v) / (150 + v)] = t + C

-(1/0.18) ln [(150 - v) / (150 + v)] = t + C

At t = 0 we have v = 0.

Thus, C = 0

Thus, -(1/0.18) ln [(150 - v) / (150 + v)] = t

a)

At max. velocity, net acceleration = 0.

Thus, 13.5 - 0.0006*v^2 = 0

v = 150 ft/s

b)

0.9*150 = 135 ft/s

-(1/0.18) ln [(150 - 135) / (150 + 135)] = t

t = 16.358 s

c)

-(1/0.18) ln [(150 - v) / (150 + v)] = t

(150 - v) / (150 + v) = e^(-0.18*t)

1 - (150 - v) / (150 + v) = 1 - e^(-0.18*t)

2v/(150+v) = 1 - e^(-0.18*t)

(150+v) / v = 2 / [1 - e^(-0.18*t)]

150/v + 1 = 2 / [1 - e^(-0.18*t)]

150/v = 2 / [1 - e^(-0.18*t)] - 1

150/v = [1 + e^(-0.18*t)] / [1 - e^(-0.18*t)]

v = 150*[1 - e^(-0.18*t)] / [1 + e^(-0.18*t)]

Since Tanh t = [1 - e^(-2*t)] / [1 + e^(-2*t)], we get

v = 150 Tanh (t*(0.18/2))

v = 150 Tanh (t*0.09)

v = dx/dt

dx = 150 Tanh (t*0.09) dt

Integrating it, x = (150*0.09)* ln Cosh (t*0.09) + Constant

At t = 0 we have x = 0.

Thus, we get Constant = 0

Thus, x = (150*0.09)* ln Cosh (t*0.09)

At t = 16.358 s, we get x = (150*0.09)* ln Cosh (16.358*0.09) = 11.21 ft