Question

8. 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 (a) 1.5 km (b)2 km (c) 2.5 km (d) 1.2 km​

Answer 1

$\tt\large\underline\purple{Given:-}$

$\tt{\implies Final\: velocity\:(v) = 90km/h}$

$\tt{\implies Initial\:velocity\:(u) = 0m/s}$

$\tt{\implies Time\: taken\:(t) = 2\:min}$

$\tt\large\underline\purple{To\: Find:-}$

$\tt{\implies Distance\: covered\:by\: train\:(s)=?}$

$\tt\large\underline\purple{Solution:-}$

To calculate the distance covered by the train ,at first we have to convert the unit of final velocity and time. Then by Applying formula to calculate the distance.

$\tt{\implies Final\: velocity=90\times\:\dfrac{5}{18}=25m/s}$

$\tt{\implies Time\: taken=2\times\:60=120\:sec}$

By using first equation of motion :-

$\tt{\implies v = u + at}$

$\tt{\implies 25 = 0 + a(120)}$

$\tt{\implies 25 = 120a}$

$\tt{\implies a =\dfrac{25}{120}m/s}$

By using 2nd equation of motion :-

$\tt{\implies s = ut + 1/2*at^2}$

$\tt{\implies s = 0(120) + 1/2\times\:\dfrac{25}{120}\times\:(120)^2}$

$\tt{\implies s = 0 + 1/2\times\:25\times\:120}$

$\tt{\implies s = 25\times\:60}$

$\tt{\implies s = 1500m}$

$\tt{\implies s = \dfrac{1500}{1000}=1.5km}$

$\bf\small\underline{Hence,\:distance\: covered\:by\: train\:(s)=1.5km}$

Answer 2

Answer:

1.5 km

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(  * 120 *120)

 = 0 + (25*120)

 = 25*60

 =  1500 m