Question

A bus accelerates at 1 m/sec² from an initial velocity of 4 m/sec for 10 sec the distance moved in that time will be ?

Answer 1

210m

by 3rd equation of motion

s=ut+1/2at^2

u=initial velocity

a=acceleration

t=time

Answer 2

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\textsf{ \to A bus accelerates from \sf 1m/s^2 .}\\\textsf{ \to It travelled for 10 seconds.}\\\textsf{ \to It had a Initial velocity of 4m/sec .}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Find :- }}}$

$\textsf{ \to The distance travelled during that time .}$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

Here we are given Initial velocity , Acceleration and time , so here we can use second equation of motion ,

$\underline{\underline{\purple{\boldsymbol{ Using\ second \ equation\ of\ motion :- }}}}$

$\sf:\implies \pink{ s = ut + \dfrac{1}{2} at^2}\\\\\sf:\implies s = (4m/s)(10s) + \dfrac{1}{2}\times 1m/s^2\times (10s)^2 \\\\\sf:\implies s = 40 m + \dfrac{1}{2}\times 100m\ \\\\\sf:\implies s = 40m + 50m \\\\\sf:\implies \boxed{\pink{\mathfrak{ Distance = 90m }}}$

$\underline{\underline{\blue{\sf \therefore Hence \ the \ Distance \ travelled \ is\ \textsf{\textbf{ 90m}}.}}}$

$\rule{200}2$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: More \ To \ Know:- }}}$

Apart from second equation of motion which is s= ut + ½at² , there are three other equations of motion , which are ,

$\bullet\ \large\underline{\orange{\tt First \ equation\ of \ motion :- }}$

The first equation of motion is v = u + at which establishes the relationship between final Velocity, initial Velocity, acceleration and time. This equation can easily be decuced from the formula of acceleration.

$\qquad\qquad\boxed{\red{\sf v = u + at }}$

$\rule{200}1$

$\bullet\ \large\underline{\orange{\tt Third \ equation\ of \ motion :- }}$

The third equation of motion is v² = u² + 2as which establishes the relationship between final Velocity, initial Velocity, acceleration and Distance . This can be derived with the help of first equation of motion .

$\qquad\qquad\boxed{\red{\sf v^2 = u^2 + 2as }}$

$\rule{200}1$

$\bullet\ \large\underline{\orange{\tt Distance\ travelled\ in \ nth\ second :- }}$

We can find the distance travelled in nth second by using the formula , s_n = u + ½ a[ 2n -1 ] , where n is the nth second. The distance travelled in the nth second means the distance travelled by the body in ( n-1) seconds to n seconds .

$\qquad\qquad\boxed{\red{\sf S_n = u + \dfrac{1}{2}a[ 2n-1] }}$

Note:-

• Here we are only talking about the object travelling in a straight line.