Question

A particle is moving with velocity of $4m{s}^{-1}$ along positive x-direction, an acceleration of $1m{s}^{-2}$ is acted upon the particle along negative x-direction. Find the distance travelled by the particle in $10s$ .

• A. $10m$
• B. $26m$
• C. $16m$
• D. $8m$

Answer 1

The correct option is B

$26m$

Velocity:

1. It can be defined as the rate of change of the object’s position with respect to a frame of reference and time.
2. It is a vector quantity because it has both magnitude and direction.

Acceleration:

1. It is a vector quantity because it has both magnitude and direction.
2. It is equal to the second derivative of position with respect to time.
3. It is equal to the first derivative of velocity with respect to time.

Step 1: Given data

The initial velocity of the particle, $u=4m{s}^{-1}$.

Acceleration applied to the particle, $a=-1m{s}^{-2}$.

Step 2: Find the time after which the particle will change its direction of motion.

As acceleration is acting upon the particle in a direction opposite to the direction of motion that's why after some time the velocity of the particle will become zero and the particle will change its direction of motion.

Firstly, find the time at which the velocity of the particle will become zero.

As we know the first equation of motion is $v=u+at$,

On putting the given data in the above equation we get:

$0=4+(-1)t$

$t=4s$

Step 3: Find the distance traveled by the particle in $\mathbf{10}\mathbf{}\mathit{s}$

As we know the second equation of motion is $s=ut+\frac{1}{2}a{t}^2$,

Therefore, the distance traveled in $10s$ is $s=(Dis\tan cetraveledin4s)+(Dis\tan cetraveledin6s)$.

Now put all the given data in the above equation to calculate the distance $s$.

$\Rightarrow s=\left[4\times 4+\frac{1}{2}\times (-1)\times {\left(4\right)}^2\right]+\left[0\times 6+\frac{1}{2}\times \left(1\right)\times {\left(6\right)}^2\right]$

$\Rightarrow s=\left(16-8\right)+\left(18\right)$

$\Rightarrow s=26m$

The distance traveled by the particle in $10s$ is $\mathbf{26}\mathbf{}\mathit{m}$. Hence, option B is correct.