Question

a car moving with constant acceleration covers the distance between two points 0.8km apart in 75sec . Its speed as it passes the second point is 20 m/s FIND:

a)What is its speed at the first point?

b)What is its acceleration?

c) At what prior distancefrom the first point was the car at rest?

c?At what prior distance fromthe first

Answer 1

(a)

we know that

v = u + at

or

a = (v - u) / t (1)

and

s = ut + (1/2)at²

or

a = 2(s - ut) / t² (2)

by equating (1) and (2), we get

(v - u) / t = 2(s - ut) / t² (3)

now, here

t = 75 secs

s = 0.8 km = 800 m

v = 20 m/s

so, by putting these values in (3), we get

(20 - u) / 75 = 2.(800 - 75u) / 75²

or

by further solving , we get

112500 - 5625u = (1600 - 150u).75

or

5625u = 7500

thus, the initial speed will be

u = 1.33 m/s

(b)

so, the acceleration will be

a = (v - u) / t

or

a = (20 - 1.33) / 75

thus, the acceleration will be

a = 0.248 m/s²

(c)

let the distance be s' and the initial velocity be u'

we shall use the following relation

u² - u'² = 2as'

or

s' = (u² - u'²) / 2a

now, in this case the initial velocity will be zero and the acceleration will remain the same, so

u' = 0

a = 0.248 m/s²

and u = 1.33 m/s

thus, we get

s' = 1.33² / (2x0.248)

so, the initial distance will be

s' = 3.56 m