Question

a train is moving at the speed 90km /h.When the brakes are applied the retardation produced is 0.5m/s^2.Find the velocity after 10 seconds and the time taken to come to rest​

Answer 1

Given :

➳ Initial velocity of train = 90kmph

➳ Retardation due to brakes = 0.5m/s²

To Find :

⟶ Velocity of train after 10s.

⟶ Time taken by train to come at rest.

Conversion :

➢ 1 km/h = 5/18 m/s

➢ 90 km/h = 90 × 5/18 = 25m/s

SoluTion :

✴ Velocity of train after 10s :

$:\implies\sf\:v=u+at$

$:\implies\sf\:v=25+(-0.5)(10)$

$:\implies\sf\:v=25-5$

$:\implies\boxed{\bf{v=20\:ms^{-1}}}$

✴ Time taken by train to come at rest :

$:\implies\sf\:v=u+at$

$:\implies\sf\:0=25+(-0.5)t$

$:\implies\sf\:-0.5t=25$

$:\implies\sf\:t=\dfrac{25}{0.5}$

$:\implies\boxed{\bf{t=50\:s}}$

[Note : Negative sign of a shows retardation.]

Answer 2

Given ,

• Initial speed (u) = 90 km/hr or 25 m/s
• Retardation (a) = -0.5 m/s²
• Time (t) = 10 sec

We know that ,

The Newton's first equation of motion is given by :

$\boxed{ \sf{v = u + at}}$

Thus ,

v = 25 + (-0.5) × 10

v = 25 - 5

v = 20 m/s

Therefore ,

• The velocity of train after 10 sec will be 20 m/s

Now , at v = 0 m/s

0 = 25 + (-0.5) × t

t = -25/(-0.5)

t = 50 sec

Therefore ,

• The time taken by a train to come at rest is 50 sec