Question

A 1100 kg car starts from rest on a horizontal road and gains a speed of 89 km/h in 35 s. (a) What is its kinetic energy at the end of the 35 s? (b) What is the average power required of the car during the 35 s interval? (c) What is the instantaneous power at the end of the 35 s interval, assuming that the acceleration is constant?

Answer 1

Answer: The answers are given below;

a) 336093,12

b) 9602,66

c) 9602,66

Step-by-step explanation:

First let's start making 89 km/h to m/s. To do this;

`(89 . 1000)/(3600) = 24,72 m/s`

a ) Now let's find its kinetic energy; As we know kinetic energy formula is
`(1)/(2) mv^(2)`

Car's kinetic energy after 35 seconds is:

`(1)/(2) .1100.24,72^(2) = 336093,12 kg(m^(2) )/(s^(2) )`

b ) Because the road is horizontal it is a fact that W(work) = Δ
`E_(k)`

Therefore the average power required is;

P(Power) =
`(W(Work))/(t(time))` and;

`P_(avg) = (336093,12)/(35) = 9602,66[W/s]`

c ) Instantaneous power can be calculated as:

`P_(ins)=F.V`

Now, since we assume that acceleration is constant, we can use the formula x = V.t to find x(distance). Then we can find F as we know W=F.x

`x = 24,72.35=865,2` and

`W=336093,12 = F.865,2` →
`F = 388,457`

therefore the instantaneous power at the end of 35 seconds is:

`P = 388,457. 24,72 = 9602,66[W/s]`