Question

A jeep starts from the state of rest.If it's velocity becomes 60 km/hr in 5 minutes,what is the acceleration of the jeep?Also calculate the distance covered by the jeep.

Answer 1

Answer:

acceleration= velocity/time

60km/hr = 60 km /60 mins = 1km/min so accn = 1km/5= 1000m/5 = 200m/m

v²= u²+2as

(1000)² = 0²+ 2 x 200 x s

(1000)²/400= s

2500m

Answer 2

Since the jeep starts from rest

initial velocity $u=0$

final velocity $v=60km/hr$

             $v=60\times \frac{5}{18} m/sec$

             $v=16.6m/sec$

time $t=5minutes\ or \ 300 seconds$

acceleration $a=\frac{change\ in\ velocity}{change\ in\ time}$

                     $a=\frac{v-u}{t}$

                     $a=\frac{16.6}{300}$

                  $a=0.05\frac{m}{sec^{2} }$

distance covered by the jeep will be

            $D=16.6\times 300$

          $D=4980m \ or \ 4.9km$

So the distance covered by the jeep will be $4.9km$

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