Question

A train starting from rest attains a velocity of 90 Km/h in 2 min, then the distance travelled by the train for attaining this velocity is * 2 points

Answer 1

Answer:

1.5 km

Explanation:

here u is the initial velocity, v is the final velocity,  t is the total time taken, a is the acceleration, s is the distance traveled.

u = 0 km/hr = 0 m/s

v = 90 km/hr = 25 m/s

t = 2 minutes = 120 sec

a = (v - u)/t

=  25/120 m/s²

Using second equation of motion,

s = u*t + 1/2(at²)

  = 0× 120 + 1/2( $\frac{25}{120}$ * 120 *120)

  = 0 + $\frac{1}{2}$(25*120)

  = 25*60

  =  1500 m

Answer 2

Answer:

1500m

Explanation:

given,

final velocity = 90km/hr= 90*(5/18)= 25m/s.

time taken by train to attain that velocity is 2min,

i.e,120secs.

we can find acceleration by using,

a=(v-u)/t,

=>a= (25-0)/120= 5/24m/s^2.

then substitute a in S = Ut+1/2(at^2),

to find the distance,

S = 0+ 1/2[(5/24)*(120*120)],

=>S = 1500m.

Hope it helps u...