Question

A train accelerated from 10km/hr to 40km/hr in 2 min. How much distance does it cover in this period ? Assume that the tracks are straight .

Answer 1

Answer:

The distance covered by the train in 2 mins is 0.83 kilometers / 830 meters.

Explanation:

In the question,

Initial speed($u$) = 10km/hr

Final speed ($v$) = 40 km/hr

Time($t$)= 2 mins = 1/60 hours= 1/30 hours

( The SI unit of time has to be changed to hours since the speeds are given in kilometers/hours)

We know that,

                $a=\frac{v-u}{t}$

Substituting the given values,

                 $a$ = $\frac{40-10}{1/30}$

                 $a=$  $900 kms/hrs$

According to the equation,

$s=ut+\frac{1}{2}at^{2}$

where $s$ stands for the distance covered.

Substituting the given values in the above equation,

        $s$ =  $10*\frac{1}{30} + \frac{1}{2}*900*1/30^{2}$

           = 0.33+0.5

           = 0.83 kms.

The distance covered by the train is 0.83 kilometers.

                       

                                 

Answer 2

Answer:

The distance covered by the train in 2 mins is 0.83 kilometers / 830 meters.

Explanation:

In the question,

Initial speed(uu ) = 10km/hr

Final speed (vv ) = 40 km/hr

Time(tt )= 2 mins = 1/60 hours= 1/30 hours

( The SI unit of time has to be changed to hours since the speeds are given in kilometers/hours)

We know that,

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

t

v−u

Substituting the given values,

aa = \frac{40-10}{1/30}

1/30

40−10

a=a= 900 kms/hrs900kms/hrs

According to the equation,

s=ut+\frac{1}{2}at^{2}s=ut+

2

1

at

2

where ss stands for the distance covered.

Substituting the given values in the above equation,

ss = 10*\frac{1}{30} + \frac{1}{2}*900*1/30^{2}10∗

30

1

+

2

1

∗900∗1/30

2

= 0.33+0.5

= 0.83 kms.

The distance covered by the train is 0.83 kilometers.