Question

) A jeep starts from the state of rest. If its velocity becomes 60 km/hr in 5 minutes, (i) What is the acceleration of the jeep? (ii) What is the distance covered by the jeep? (Ans: 0.056 m/s", 2500 m) ​

Answer 1

Answer:

Ans: a=0.056ms^-2  S=2500m

Explanation:

vi=0ms^-1

vf=60km/hr= $\frac{60*1000}{3600}$=16.6ms-1

a=$\frac{vf-vi}{t}$

 =$\frac{16.6-o}{5*60}$

a=0.056ms^-2

S=?

S=vit+$\frac{1}{2}$$at^{2}$

S=0+$\frac{1}{2}$*0.056*$(300)^{2}$

S=2500m