(a) Let's convert the final speed of the car in m/s:

The kinetic energy of the car at t=19 s is

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:

But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into

(c) The instantaneous power is given by

where F is the force exerted by the engine, equal to F=ma.
So we need to find the acceleration first:

And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
Answer:
Magnitude of force on wheel B is 4 N
Explanation:
Given that

For wheel A
m= 1 kg
d= 1 m,r= 0.5 m
F=1 N
We know that
T= F x r
T=1 x 0.5 N.m
T= 0.5 N.m
T= I α
Where I is the moment of inertia and α is the angular acceleration


T= I α
0.5= 0.25 α

For Wheel B
m= 1 kg
d= 2 m,r=1 m


Given that angular acceleration is same for both the wheel

T= I α
T= 1 x 2
T= 2 N.m
Lets force on wheel is F then
T = F x r
2 = F x 1
So F= 2 N
Magnitude of force on wheel B is 2 N
i sorry i thought of geocentric as something else it appears that the earth was the center
It was be at least 67 cm/yr for the North Atlantic