Question

A particle is moving in a straight line with initial velocity and uniform acceleration a. If the sum of the distance travelled in ${\mathrm{t}}^{\mathrm{th}}$ and ${(\mathrm{t}+1)}^{\mathrm{th}}$ seconds is $100\mathrm{cm}$, then its velocity after $\mathrm{t}$ seconds, in $\mathrm{cm}/\mathrm{s}$, is

Answer 1

Step 1. Given data

The sum of the distance travelled in ${\mathrm{t}}^{\mathrm{th}}$ and ${(\mathrm{t}+1)}^{\mathrm{th}}$ seconds is $100\mathrm{cm}$.

We have to find the velocity.

Step 2. Concept used.

The distance travelled in ${\mathrm{n}}^{\mathrm{th}}$ second is,

${\mathrm{S}}_{\mathrm{n}}=\mathrm{u}+\frac{1}{2}\left(2\mathrm{n}-1\right)\mathrm{a}$ ----------- $\left(1\right)$

Here, $\mathrm{S}$ is the distance travelled time, $\mathrm{u}$ is the initial velocity, and $\mathrm{a}$ is acceleration of the body in motion.

Step 3. Calculate the velocity.

The distance travelled in ${\mathrm{t}}^{\mathrm{th}}$ and ${(\mathrm{t}+1)}^{\mathrm{th}}$ second is $100\mathrm{cm}$.

Now we will find the sum of distances.

By using the expression $\left(1\right)$, the distance travelled in ${\mathrm{t}}^{\mathrm{th}}$ is,

${\mathrm{S}}_{\mathrm{t}}=\mathrm{u}+\frac{1}{2}\left(2\mathrm{t}-1\right)\mathrm{a}$ ------ $\left(2\right)$

And the distance travelled in ${(\mathrm{t}+1)}^{\mathrm{th}}$ is,

${\mathrm{S}}_{\mathrm{t}+1}=\mathrm{u}+\frac{1}{2}\left(2\left(\mathrm{t}+1\right)-1\right)\mathrm{a}$

$=\mathrm{u}+\frac{1}{2}\left(2\mathrm{t}+1\right)\mathrm{a}$---------- $\left(3\right)$

According to the given, the total distance travelled in ${\mathrm{t}}^{\mathrm{th}}$ and ${(\mathrm{t}+1)}^{\mathrm{th}}$ second is $100\mathrm{cm}$, so, we have to add expression $\left(1\right)$ and $\left(2\right)$, we get,

${\mathrm{S}}_{\mathrm{t}}+{\mathrm{S}}_{\mathrm{t}+1}=100$

Substitute the values of expression $\left(1\right)$ and $\left(2\right)$ in above expression, we get,

$\mathrm{u}+\frac{1}{2}\left(2\mathrm{t}-1\right)\mathrm{a}+\mathrm{u}+\frac{1}{2}\left(2\mathrm{t}+1\right)\mathrm{a}=100$

$2\left(\mathrm{u}+\mathrm{at}\right)=100$

$\left(\mathrm{u}+\mathrm{at}\right)=\frac{100}{2}$

$\left(\mathrm{u}+\mathrm{at}\right)=50$ ------$\left(4\right)$

Now, from the first equation of motion the velocity of any particle after time $\mathrm{t}$, if it moves with acceleration $\mathrm{a}$ is,

$\mathrm{v}=\mathrm{u}+\mathrm{at}$ -------- $\left(5\right)$

Now, its clear that $\mathrm{u}+\mathrm{at}$ is similar in expression $\left(5\right)$ as well as $\left(4\right)$, so substitute the value of $\mathrm{u}+\mathrm{at}$ from expression $\left(4\right)$ in $\left(5\right)$, we get

$\mathrm{v}=\mathrm{u}+\mathrm{at}$

$\mathrm{v}=50\mathrm{cm}/\mathrm{s}$

Hence, a particle is moving in a straight line with initial velocity and uniform acceleration a. If the sum of the distance travelled in ${\mathrm{t}}^{\mathrm{th}}$ and ${(\mathrm{t}+1)}^{\mathrm{th}}$ seconds is $100\mathrm{cm}$, then its velocity after $\mathrm{t}$ seconds, in $\mathrm{cm}/\mathrm{s}$, is $50\mathrm{cm}/\mathrm{s}$.