Question

A train is traveling at a speed of 90 km /h-1 beakes are applied so to produce a uniform acceleration of -0.5 m/s-2 find how far train will go before it is brought to rest

Answer 1

Given :

▪ Initial speed = 90kmph = 25mps

▪ Acceleration of train = -0.5m/s$^2$

(Negative sign shows retardation.)

To Find :

▪ Distance covered by train before it is brought to rest.

Concept :

✏ This question is completely based on concept of stopping distance.

✏ Since, acceleraration has said to be constant throughout the motion we can easily apply equation of kinematics to solve this type of questions.

Formula :

✒ Third equation of kinematics :

$\bigstar\underline{\boxed{\bf{\pink{v^2-u^2=2as}}}}$

• v denotes final speed
• u denotes initial speed
• a denotes acceleration
• s denotes distance

Calculation :

→ v$^2$-u$^2$ = 2as

→ (0)^2 - (25)^2 = 2(-0.5)s

→ -625 = (-1)s

→ $\underline{\boxed{\bf{\purple{s=625\:m}}}}$

Answer 2

$\rule{200}3$

$\huge\tt{GIVEN:}$

• A train travelling at the speed of 90 km/hr - 1, beakes are applied to produce a uniform acceleration of -0.5 m/s -2

$\rule{200}3$

$\huge\tt{TO~FIND:}$

• How far the train would go before it is on rest.

$\rule{200}3$

$\huge\tt{CONCEPT~USED:}$

We know the third equation of kinematics,

• v² - u² = 2as

Where , v = final speed , u = initial speed , a = acceleration , s = Distance

$\rule{200}3$

$\huge\tt{SOLUTION:}$

So, if we put these equation in this case,

➠(0)² - (25)² = 2(-0.5)s

➠- 625 = (-1)s (cancelling negative signs by each other)

➠625 = s

${\boxed{\fbox{\fbox{\red{DISTANCE~=~625~Meter}}}}}$

$\rule{200}3$