Question

5. A train accelerated from 10km/hr to 40km/hr in 2 minutes. How much distance does it cover in this period? Assume that the tracks are straight?
6. A train starts from rest and accelerate

Answer 1

$\large\underline{ \underline{ \sf \maltese{ \: Correct \: Question:- }}}$

A train accelerated from 10km/hr to 40km/hr in 2 minutes. How much distance does it cover in  this period?

Assume that the tracks are straight.

$\large\underline{ \underline{ \sf \maltese{ \: Given:- }}}$

$\boxed{\sf{u=10km/h=2.78m/s}}$

$\boxed{\sf{v=40km/h=11.11m/s}}$

$\boxed{\sf{t=2minutes=120s}}$

$\large\underline{ \underline{ \sf \maltese{ \:To\:Find:- }}}$

Distance covered by the train in the interval of time.

$\large\underline{ \underline{ \sf \maltese{ \: Solution:- }}}$

We know that,$\boxed{\sf{v=u+at}}$

$\boxed{\sf{a=}{\dfrac{v-u}{t}}}$

$\sf{a=}{\dfrac{11.11-2.78}{120}}$

$\sf{a=0.07ms^{-2}}$

Also, $\boxed{\sf{s=ut+{\dfrac{1}{2}}at^2}}$

$\sf{\implies}{s=2.78 \times 120+{\dfrac{1}{2}}\times 0.07 \times 120\times 120}$

$\sf{\implies}{s=333.6+504}$

$\boxed{\sf{\implies}{\red{s=837.6m}}}$

$\large\underline{ \underline{ \sf \maltese{ \: More\:to\:know:- }}}$

Acceleration:

Acceleration is the rate of change of the velocity of an object with respect to time. Accelerations are vector quantities (in that they have magnitude and direction).

Mathematically,

$\boxed{\sf{a={\dfrac{v-u}{t}}}}$

Where,

• a= Acceleration
• v= Final Velocity
• u= Initial Velocity
• t= time taken.