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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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Sagot :

rafaleo84

Answer:

See the answers below

Explanation:

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

[tex]v_{f}=v_{o}+a*t[/tex]

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

[tex]72[\frac{km}{h}]*[\frac{1000m}{1km}]*\frac{1h}{3600s} =20[m/s][/tex]

i)

Vf = final velocity = 20 [m/s]

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

a = acceleration [m/s²]

t = time = 4 [s]

[tex]20 = 0+a*4\\a=20/4\\a = 5[m/s^{2}][/tex]

ii)

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

[tex]v_{f}^{2}=v_{o}^{2} +2*a*x[/tex]

where:

x = distance covered [m]

[tex]20^{2} =0 +2*5*x\\400=10*x\\x = 40 [m][/tex]

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