Question

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.

Answer 1

When he weight of the car is 8.55 x
`10^(3)` N then power = 314.012 KW

When he weight of the car is 1.10 x
`10^(4)` N then power = 43.76 KW

Step-by-step explanation:

Given that

Initial velocity
`V_(1)` = 0

Final velocity
`V_(2)` = 24.8
`(m)/(s)`

Time = 7.88 sec

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

P =
`(change \ in \ kinetic \ energy)/(time)`

Change in kinetic energy Δ K.E =
`(1)/(2) m (V_(2)^(2) - V_(1)^(2) )`

Since Initial velocity
`V_(1)` = 0

⇒ Δ K.E =
`(1)/(2) m V_(2) ^(2)`

⇒ Power P =
`(1)/(2) (m)/(t) V_(2) ^(2)`

⇒ Power P =
`(1)/(2) (W)/(g\ t) V_(2) ^(2)` -------- (1)

(a). The weight of the car is 8.55 x
`10^(3)` N = 8550 N

Put all the values in above formula

So power P =
`(1)/(2) (8550)/((9.81)\ (7.88)) (24.8) ^(2)`

P = 314.012 KW

(b). The weight of the car is 1.10 x
`10^(4)` N = 11000 N

Put all the values in equation (1) we get

P =
`(1)/(2) (11000)/((9.81)\ (7.88)) (24.8) ^(2)`

P = 43.76 KW