Question

The driver of a car, which is initially at rest at the top A of the grade, releases the brakes and coasts down the grade with an acceleration in feet per second squared given by a = 3.22-0.004v2 , where v is the velocity in feet per second. Determine the velocity vB at the bottom B of the grade.

Answer 1

Answer:

28.3 ft/s

Explanation:

We are given that

Initial velocity of a car=$u=0$

Acceleration of the car=$a=3.22-0.004v^2$

We have to find the velocity $v_B$ at the bottom of the grade.

Distance covered by car=600 ft

We know that

Acceleration=a=$\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=v\frac{dv}{ds}$

$v=\frac{ds}{dt}$

$ds=\frac{vdv}{a}$

Taking integration on both side and taking limit of s from 0 to 600 and limit of v from 0 to $v_B$

$\int_{0}^{600}ds=\int_{0}^{v_B}\frac{vdv}{3.22-0.004v^2}$

$[S]^{600}_{0}=-\frac{1}{0.008}[ln(3.22-0.004v^2)]^{v_B}_{0}$

Using substitution method  and $\int f(x)dx=F(b)-F(a)$

$600-0=-\frac{1}{0.008}[ln(3.22-0.004v^2_B)-ln(3.22)]$

$600=-\frac{1}{0.008}(ln(\frac{3.22-0.04v_B}{3.22}))$

$ln(m)-ln(n)=ln(\frac{m}{n})$

$600\times 0.008=-ln(\frac{3.22-0.04v_B}{3.22})$

$-4.8=ln(\frac{3.22-0.04v_B}{3.22})$

$\frac{3.22-0.04v_B}{3.22}=e^{-4.8}$

$ln x=y\implies x=e^y$

$3.22-0.004v^2_B=3.22e^{-4.8}$

$3.22-0.004v^2_B=0.0265$

$3.22-0.0265=0.004v^2_B$

$0.004v^2_B=3.22-0.0265=3.1935$

$v_B=\sqrt{\frac{3.1935}{0.004}}=28.3 ft/s$

Hence, the velocity $v_B$ at the bottom of the grade=28.3 ft/s

Answer 2

The velocity of the car at the bottom of the grade is 28.37 ft/s.

What is acceleration?

Acceleration is the rate of change of velocity with time.

Motion of the car from top to bottom of the grade

The velocity of the car increases as the car moves downwards and it will be maximum when the car reaches equilibrium position. At equilibrium position (bottom of the grade) the acceleration will be zero but the velocity will be maximum.

a = 3.22 - 0.004v²

0 = 3.22 - 0.004v²

0.004v² = 3.22

v² = 3.22/0.004

v² = 805

v = √805

v = 28.37 ft/s

Thus, the velocity of the car at the bottom of the grade is 28.37 ft/s.

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