Question

a train starting rest attains a velocity of 72 km/h in 5min (a) speed and distance​

Answer 1

Explanation:

Lets, firstly find the acceleration:-

u=0, v=72km/h Or 20m/s

t=5 min Or 300 seconds

a = v-u/t = 20-0/300 = 20/300 = 1/15m/s²

Distance=

s=ut +(1/2)at²

=0+(1/2)(1/15)(300)(300)

= 3 km Or 3000 m

Speed = Distance/Time

3000/300 = 30 seconds

Answer 2

$\large\bold\green{☑Verified Answer:-}$

Given

Initial velocity = u = 0m/s

Final velocity = v = 72 km/h

Time = 5 min

We will convert mins into hrs.

$\frac{5}{60} hrs = \frac{1}{12} hrs$

$a = \frac{72 - 0}{ \frac{1}{2} }$

a = 72 × 12

= 864 km/h²

Hence the acceleration is 864 km/h²

Now, the distance travelled by train =

$0 \times \frac{1}{12} + \frac{1}{2} (864) + (\frac{1}{12} ) {}^{2}$

$\frac{1}{2} \times 864 \times \frac{1}{144}$

= 3 km

Hence, the distance travelled by train in 3 km.