Question

a jeep starts from the state of rest. If its velocity becomes 60km/hrin 5minutes,what is the acceleration of the jeep? What is the distance covered by the jeep?

Answer 1

Answer:

Acceleration is 0.056m/sq m

Distance 2500m

Explanation:

• initial velocity is 0m/sec

• Final velocity is 60Km/hr or 16.667 m/sec

$a = \frac{v - u}{t} = \frac{16.667}{300}$

$= 0.056 \: m \: {sec}^{ - 2}$

1. Distance

• ${v}^{2} - {u}^{2} = 2as$
• Solving we get
• s=2500m

Answer 2

Answer:

acceleration=0.056

distance covered=2500m

Explanation:

i> final velocity(v)=60km/hr

                         =60*1000m/60*60s

                         =50m/3s =16.67m/s

initial velocity(u)=0

time(t)=5min

          =5*60s

           =300s

now  

       a=v-u/t

          =16.67m/300s

          =0.056m/s^2

                                                                                                                                   

ii> 60km/hr

=60km/60min

=1km/min

so according to ques,

=1km/5min

=1000m/5

=200m/m

now,

v²= u²+2as

(1000)² = 0²+ 2 x 200 x s

(1000)²/400= s

2500m

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