Question

A particle moving with uniform acceleration travels a distance of 20 m in 5th second and 25 m in 7th second. The distance travelled in 11th second will be​

Answer 1

Given:

The particle moves with uniform acceleration

Distance travelled in $5^{th}$ second is $d_{5} =20m$

Distance travelled in $7^{th}$ second is $d_{7} =25m$

To Find:

The distance travelled in $11^{th}$ second

Step by Step Explanation:

Acceleration is given by $a=\frac{dv}{dt}$

where, $dv$ is change in velocity and $dt$ is change in time.

Let $t_{5}, t_{7}, t_{11}$ represent the time and $v_{5}, v_{7}, v_{11}$ represent the velocity at $5^{th} ,7^{th}, 11^{th}$ second respectively.

Determine the Change in Velocity:

Velocity of the particle at $5^{th}$ second  $v_{5} =20m/s$

Velocity of the particle at $7^{th}$ second $v_{7} =25 m/s$

Change in velocity $dv=v_{7} -v_{5}$

$=25-20$

⇒$dv=5m/s$

Determine the Change in Time:

Change in time $dt=t_{7} -t_{5}$

$=7-5$

⇒$dt=2s$

Determine the acceleration:

Acceleration $a=\frac{dv}{dt}$

$=\frac{5}{2}$

⇒$a=2.5m/s^{2}$

Determine the distance travelled in $11^{th}$ second:

Consider $v_{11}$ as final velocity $v$ and $v_{7}$ as initial velocity $u$

Let $t$ be the time duration between $t_{7}$ and $t_{11}$

∴ $t=4s$

From the equation of motion, $v=u+(a\times t)$ we get;

$v_{11} =v_{7} +(a\times t )$

$=25+(2.5 \times 4)$

⇒$v_{11} =35m/s$

∵ Velocity is distance travelled by time, the distance travelled in $11^{th}$ second $= 35m$

Answer:

$35m$