Question

A jeep starts from rest. If its velocity becomes 72 km/hr in 4 seconds.

i. What is the acceleration of the jeep ?
[Ans: 5 m/s]
ii. What is the distance covered by jeep?
[Ans : 10 m]
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Answer 1

See the answers below

Step-by-step explanation:

To solve this problem we must use the following equation of kinematics.

`v_(f)=v_(o)+a*t`

But first we must convert the velocity from kilometers per hour to meters per second.

`72[(km)/(h)]*[(1000m)/(1km)]*(1h)/(3600s) =20[m/s]`

i)

Vf = final velocity = 20 [m/s]

Vo = initial velocity = 0 (stars from the rest)

a = acceleration [m/s²]

t = time = 4 [s]

`20 = 0+a*4\\a=20/4\\a = 5[m/s^(2)]`

ii)

Now we can calculate the distance with the following equation of the kinematics.

`v_(f)^(2)=v_(o)^(2) +2*a*x`

where:

x = distance covered [m]

`20^(2) =0 +2*5*x\\400=10*x\\x = 40 [m]`