Question

Train starts from rest and accelerate uniformly at the rate of 5 m/s2 for 5 sec. calculate i. the velocity of train in 5 sec. ii. the distance covered by the train in this time.

Answer 1

Given :

▪ Initial velocity of train = zero

▪ Acceleration of train = 5m/s²

▪ Time interval = 5s

To Find :

▪ Final velocity and distance travelled by train after the given intetval of time.

Concept :

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

✴ First eqauation of kinematics :

$\bigstar\:\underline{\boxed{\bf{\red{v=u+at}}}}$

✴ Third equation of kinematics :

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

Calculation :

❄ Final velocity of train :

$\dashrightarrow\tt\:v=u+at\\ \\ \dashrightarrow\tt\:v=0+(5\times 5)\\ \\ \dashrightarrow\:\blue{\underline{\boxed{\tt v=25\:mps}}}$

❄ Distance covered by train :

$:\implies\tt\:v^2-u^2=2as\\ \\ :\implies\tt\:(25)^2-(0)^2=2(5)s\\ \\ :\implies\tt\:s=\dfrac{625}{10}\\ \\ :\implies\:\orange{\underline{\boxed{\tt s=62.5\:m}}}$

Answer 2

$\huge{\pink{\underline{\underline{Question:}}}}$

Train starts from rest and accelerate uniformly at the rate of 5 m/s 2 for 5sec.

1) Calculate velocity of train in 5sec

2)The distance covered by the train in this time

$\huge{\purple{\underline{\underline{Answer:}}}}$

Given,

Initial velocity of a train(u) = Zero

Acceleration of train(a) = 5m/s^2

Time (t)= 5s

We know that,

First kinematic equation = V=u+at

V= ?

U= 0

a=5

t=5

V = 0+(5×5)

V=5×5

V=25mps

so,

Final velocity of a train is 25 mps.

---------------------------------------------------

Third kinematic equation :

${v}^{2} - {u}^{2} = 2as$

${25}^{2} - {0}^{2} = 2(5)s$

$625 - 0 = 10s$

$s = \frac{625}{10}$

S =62.5m

So, The distance covered by train = 62.5m.