Answer to Question #188564 in Mechanics | Relativity for Stacey Chua

Question

What is the final velocity of a 1200 kilogram car starting from rest after 50 meters when 4500 Newton's of force is applied?

Answer 1

According to the second Newton's law, the acceleration of the car is:

$a=\frac{F}{m}$

F=4500N is total force applied to the car and m=1200kg Is the mass of car

According to the kinematic equation, the distance covered under a constant acceleration is:

$d=v_{0}t+\frac{1}{2}at^2$

v₀=0

Since d=50m

$d=\frac{1}{2}at^2$

$t=\sqrt\frac{2d}{a}=\sqrt\frac{2dm}{F}$

The velocity of the car at this moment of time (if it starts from rest) is given by the following kinematic equation:

$v=v_{0}+at$

Substituting the expressions for aa and tt, obtain:

$v=\frac{F}{m}\sqrt\frac{2dm}{F}=\sqrt{\frac{2dF}{m}}$

Put value

$v=\sqrt{\frac{2\times250\times4500}{1200}}=6.1m/sec$

Final velocity of car is 6.1m/sec