Question

solve this numerical by using equations of motion! ​

Answer 1

Given :

▪ Initial velocity of train = 20mps

▪ Deceleration = 8m/s²

▪ Final velocity of train = zero (i.e., rest)

To Find :

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

Concept :

❇ This question is completely based on the concept of stopping distance.

❇ Since, retardation has said to be constant throughout the motion, we can easily apply equation of kinematics to solve this question.

❇ Third equation of kinematics :

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

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

Calculation :

$\dashrightarrow\sf\:v^2-u^2=2as\\ \\ \dashrightarrow\sf\:(0)^2-(20)^2=2(-8)s\\ \\ \dashrightarrow\sf\:-400=-16s\\ \\ \dashrightarrow\sf\:s=\dfrac{400}{16}\\ \\ \dashrightarrow\underline{\boxed{\bf{\purple{s=25m}}}}\:\gray{\bigstar}$

[Note : -ve sign of a shows retardation]

Answer 2

GiVeN :-

→ A train moving at 20 m/s breaks and decelerates uniformly at 8 m/s².

• Initial velocity, u = 20 m/s
• Deceleration, a = -8 m/s²

(Negative sign signifies deceleration)

• Final velocity, v = 0 m/s

(As the train comes to rest at last)

To DeTeRmInE :-

The distance covered(s) by the train before coming to rest.

AcKnOwLeDgEmEnT :-

As per the 3rd equation of motion,

$\boxed{\rm{\bullet\ v^2=u^2+2as}}$

SoLuTiOn :-

Substituting the known values in the above equation :-

$\rm{\hookrightarrow (0)^2=(20)^2+2\times-8\times s}\\\\\rm{\hookrightarrow 0=400-16s}\\\\\rm{\hookrightarrow 16s=400}\\\\\rm{\hookrightarrow s=\dfrac{400}{16} }\\\\\boxed{\boxed{\rm{\hookrightarrow s=25\ m}}}$

Therefore, the train covers a distance of 25 m before coming to a stop.