A 1500 kg car accelerates uniformly from rest to reach 10.0 m sin 5.00 s. Determine (i) the average power delivered by the engine.

1.50 * 10⁴ W

Explanation:

Given data

Mass (m) : 1500 kg

Initial speed (vi) : 0 m/s (rest)

Final speed (vf) : 10.0 m/s

Time (t) : 5.00 s

First, we will calculate the acceleration (a) of the car.

a = vf - vi / t = (10.0 m/s - 0 m/s) / 5.00 s = 2.00 m/s²

Next, we find the associated force (F) using Newton's second law of motion.

F = m. a = 1500 kg. 2.00 m/s² = 3.00 * 10³ N

Then, we find the distance (d) traveled in this time.

d = 1/2. a. t² = 1/2. 2.00 m/s². (5.00 s) ² = 25.0 m

The work (w) exerted by the engine is:

w = F. d = 3.00 * 10³ N. 25.0 m = 7.50 * 10⁴ J

Finally, the average power (P) delivered by the engine is:

P = w / t = 7.50 * 10⁴ J / 5.00 s = 1.50 * 10⁴ W

Answer:240W

Explanation: from kinematics x=a*t*t/2 so we can get acceleration:

10=a*5*5/2; a = 0.8m/s2

from newton's law we get the force that moves the car:

F=m*a, F = 1500*0.8=1200N

now we get the total work done by the force F:

W = Fd = 1200*10=12000J

The power is P=W/t:

P = 12000J / 5s = 240W = 0.32HP