Question

From a station X a train starts from rest and attains a speed of 54km/hr in 10 sec. Then by applying brakes negative accleration of -2.5 m/s^2 is produced and it stops at station Y in 6 sec. Find the distamce b/w the Station X &Y​

Answer 1

In First Case

Intial Velocity (u)=0 m/s

Final Velocity (v)=54 km/h

In m/s

=54×5/18

=15 m/s

Final Velocity=15 m/s

Time=10 sec

To find Distance Travel Between 10 Seconds

$\frac{v + u}{2} \times t \\ \\ \frac{15 + 0}{2} \times 10 \\ \\ \frac{15}{\cancel2} \times \cancel{10} \\ \\ 15 \times 5 \\ = 75 \: m$

Distance Travelled in 1st 10 seconds

=75 m

In second Case When Brake Were applied

Intial Velocity (u)=15 m/s

Final Velocity (v)=0 m/s

Retardation (r)=-2.5 m/s²

Distance (s)=??

For Finding Distance

Use Equation of Motion

v²=u²+2rs

0²=15²+2×-2.5×s

0=225-5s

0-225=-5s

-225=-5s

$s = \frac{\cancel - 225}{ \cancel- 5} \\ \\ s = \frac{225}{5} \\ \\ s = 45m$

Distance Travelled to Y station from where Brake Applied=45 m

Total Distance=75+45=120 m

$\boxed{\mathbf{Total\: Distance \:from\:X\:to\:Y=120m}}$