Question

A train accelerated from 10m/s to 40 m/s in 2 second.Distance covered by body during this period is

Answer 1

$\fcolorbox{yellow}{pink}{Answer:}$

Given data,

initial speed(u)=10 m/s

final speed (v)=40m/s

time(t)=2s

To find,

distance(d)=?

Calculations,

as we know that,

$a = \frac{v - u}{t} \\ a = \frac{40 - 10}{2} \\ a = \frac{30}{2} \\ a = 15 m/ {s}^{2}$

therefore,

${v}^{2} = {u}^{2} + 2as \\ 1600 = 100 + (2 \times 15 \times s) \\ 1500 = 30s \\ s = \frac{1500}{30} \\ s = 50m$

in the given question no condition is given for direction

hence,

displacement(s)=distance(d)

distance ,d=50m

therefore total distance covered by the train in 2s is 50m

hope it helps to you☺️