Question

A 1.50 ✕ 103-kg car starts from rest and accelerates uniformly to 17.0 m/s in 12.5 s. Assume that air resistance remains constant at 400 N during this time. (a) Find the average power developed by the engine. hp (b) Find the instantaneous power output of the engine at t = 12.5 s, just before the car stops accelerating. hp

Answer 1

The force applied by air on the car = 400 N

The deacceleration due to the air on the car = 400/(1.5 x 10³) = 0.27 m/s²

Let the car has an acceleration of a_c .

So, the net acceleration (a) of the car is (a_c-0.27) in m/s².

Given, the speed of the car becomes 17.0 m/s in 12.5 s. So,

                                    

                       or,

So, total force applied by car's engine is 1.627 x 1.5 x 10³ = 2.44 x 10³ N.

The distance travelled by the carin 12.5 seconds can be calculated as

                          

                    or,

So, workdone by the engine in 12.5 seconds is

                          

(a) So, power delivered by the engine in 12.5 seconds is

                           

(b) Instantaneous power is given by

                           

At t=12.5 seconds, the velocity (v) of the car was 17 m/s.

And force applied by the engine = 2.44 x 10³ N.

So, Instantaneous power output of the engine at 12.5 second is

                              

For any doubt please comment.