Question

27 Provided a racing car does not lose traction, the
time taken by it to race from rest through a
distance x depends mainly on engine's power P.
Distance x covered in time t is
3/2
8P 1312
(1) Vamor
3/2
(2) Jan 1312
3 m
13/2​

Answer 1

Answer:

$\sqrt{\frac{8P}{9m}} t^{3/2}$

Explanation:

1) We have,

m : mass of car

v : velocity of car at time 't'

Engine Power is constant .

Now,

$P = F.v = m\frac{dv}{dt}v\\ \\=>\int_0^t \frac{Pdt}{m}=\int_0^vvdv \\ \\ =>\frac{Pt}{m}=\frac{v^2}{2}\\ \\=> v=\sqrt{\frac{2P}{m}}t^{1/2}$

2) Then,

$\frac{dx}{dt}=\sqrt{\frac{2P}{m}}t^{1/2}\\ \\=>dx=\sqrt{\frac{2P}{m}}t^{1/2}dt\\ \\=>x= \sqrt{\frac{2P}{m}}\frac{t^{3/2}}{3/2}=\sqrt{\frac{8P}{9m}} t^{3/2}$

Above x is our required answer.

Answer 2

Answer:

hope it helps you!!!!!!!!!