Question

A particle is moving in a straight line with initial velocity and uniform acceleration a. If the sum of the distance travelled in ${{t}^{th}}$ and ${{\left( t+1 \right)}^{th}}$ seconds is 100 cm, then its velocity after $t$ seconds in${cm}/{s}\;$, is
$\left( a \right)\ 80$
$\left( b \right)\ 50$
$\left( c \right)\ 20$
$\left( d \right)\ 30$

Answer 1

Hint: In this question, using the formula of distance travelled in ${{n}^{th}}$ i.e. $Sn=u+\dfrac{1}{2}\left( 2n-1 \right)a$, we will find the distance travelled in ${{t}^{th}}$ and ${{\left( t+1 \right)}^{th}}$. Now, using the formula for relation between u, v and a i.e. $v=u+at$ we will find the velocity after t seconds.

Formula Used: $Sn=u+\dfrac{1}{2}\left( 2n-1 \right)a$

Complete step-by-step answer:
First of all, in the question it is given that sum of distance travelled in ${{t}^{th}}$ and ${{\left( t+1 \right)}^{th}}$ seconds is 100 cm we will find the sum of distances. Now, we know that the distance travelled in n seconds is given by,
$Sn=u+\dfrac{1}{2}\left( 2n-1 \right)a$ ………………………(i)
Where, $Sn$ is speed in n seconds, u is initial velocity, a is acceleration and t is time in seconds.
Now, using the expression (i) we will find the distance travelled in ${{t}^{th}}$ which is given by,
${{S}_{t}}=u+\dfrac{1}{2}\left( 2t-1 \right)a$ ………………………………(ii)
And distance travelled in ${{\left( t+1 \right)}^{th}}$ is given by,
${{S}_{t+1}}=u+\dfrac{1}{2}\left( 2\left( t+1 \right)-1 \right)a\Rightarrow u+\dfrac{1}{2}\left( 2t+1 \right)a$ …………………………………(iii)
Now, as it is given that total distance, travelled in ${{t}^{th}}$ and ${{\left( t+1 \right)}^{th}}$ seconds is 100 cm, so, we will add expression (i) and (ii) as below,
${{S}_{t}}+{{S}_{t+1}}=100$
Now, we will substitute the values of expression (i) and (ii) in above equation,
$u+\dfrac{1}{2}\left( 2t-1 \right)a+u+\dfrac{1}{2}\left( 2t+1 \right)a=100$
Now, simplifying the above equation we will get,
$\Rightarrow 2\left( u+at \right)=100$
$\Rightarrow \left( u+at \right)=\dfrac{100}{2}=50$ …………………(iv)
Now, using the relation between u, v and a, we will find the value of final velocity which is given by,
$v=u+at$ …………………….(v)
Now, we can see that $u+at$ is similar in expression (v) as well as (vi), so we will substitute the value of $u+at$ from expression (iv) in expression (v) and then we will get,
$v=u+at\Rightarrow v=50\ {cm}/{s}\;$.
Hence, we can say that in t seconds the velocity of the particle will be $50\ {cm}/{s}\;$
Thus, option (b) is correct.

Note:Newton’s formulas which provide relations between time, acceleration, speed and distance are given as, $v=u+at$, ${{v}^{2}}={{u}^{2}}+2as$ and $s=u+\dfrac{1}{2}a{{t}^{2}}$ where $u$ is initial velocity and $v$ is final velocity. Depending on the values provided we can use these formulas. Units in all the formulas remain the same i.e. in the MKS unit system.