A car accelerates uniformly from rest to 24.8 m/s in 7.88 s along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if (a) the weight of the car is 8.55 x 103 N, and (b) the weight of the car is 1.10 x 104 N.

When he weight of the car is 8.55 x N then power = 314.012 KW

When he weight of the car is 1.10 x  N then power =  43.76 KW

Explanation:

Given that

Initial velocity = 0

Final velocity = 24.8

Time = 7.88 sec

We know that power required to accelerate the car is given by

P =

Change in kinetic energy Δ K.E =

Since Initial velocity = 0

⇒ Δ K.E =

⇒ Power P =

⇒ Power P =   -------- (1)

(a). The weight of the car is 8.55 x N = 8550 N

Put all the values in above formula

So power P =

P = 314.012 KW

(b). The weight of the car is 1.10 x N = 11000 N

Put all the values in equation (1) we get

P =

P = 43.76 KW

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form