Question

A particle covers $10m$ in first $5sec$ and $10m$ in next $3sec$. Assuming constant acceleration find the initial speed acceleration and distance covered in next $2sec$.

Answer 1

Above problem is related to the motion of the body in a rectilinear. So, we can easily apply the equations of motion to solve this problem. The motion of the body in which the body travels along a straight line is called rectilinear motion. This motion is an example of translatory motion.

Formula used: - To solve rectilinear problems, we use equations of motion. There are three equations of motion, by which we can find initial velocity, final velocity, acceleration, displacement and time taken by the bodies in their travelling path. The equations of motion are following-
$v = u + at$
$S = ut + \dfrac{1}{2}a{t^2}$
${v^2} = {u^2} + 2as$
Here $u$ is the initial velocity, $v$ is the final velocity, $a$ is the acceleration produced in the body, $s$is the displacement covered by the body and $t$is the time taken by the body in travelling total displacement.

Complete step by step solution: -
In the given question, we have a particle which travels $10m$in first$5sec$.So for finding initial velocity, we can apply a second equation of motion, because we know the displacement and time.
From second equation of motion $S = ut + \dfrac{1}{2}a{t^2}$
According to the question, ${s_1} = 10m,{t_1} = 5\sec$
So, ${S_1} = u{t_1} + \dfrac{1}{2}a{t_1}^2$
Substituting the values, we get
$10 = u \times 5 + \dfrac{1}{2} \times a \times {5^2} \\\ \Rightarrow 10 = 5u + \dfrac{1}{2} \times a \times 25 \\\ \Rightarrow 10 = \dfrac{{10u + 25a}}{2} \\\ \Rightarrow 20 = 10u + 25a \\\$
Or we can write- $4 = 2u + 5a$ ……………...(i)
Now, in the next$3sec$, particles are displaced $10m$ more. So, the total displacement will be $20m$and the total time is$3 + 5 = 8\sec$. So, again by second equation of motion,
${S_2} = u{t_2} + \dfrac{1}{2}a{t_2}^2$
Assuming constant acceleration and substituting-

$${S_2} = {S_1} + 10 \\\ \Rightarrow {S_2} = 10 + 10 \\\ \Rightarrow {S_2} = 20m \\\$$

And

$${t_2} = {t_1} + 3 \\\ \Rightarrow {t_2} = 5 + 3 \\\ \Rightarrow {t_2} = 8\sec \\\$$

So,
$20 = 8u + \dfrac{1}{2} \times a \times {8^2} \\\ \Rightarrow 20 = 8u + \dfrac{1}{2} \times 64 \times a \\\ \Rightarrow 20 = 8u + 32a \\\$
Or we can write $5 = 2u + 8a$ ………………...(iii)
Now, subtracting equation (i) and equation (iii),
$5 - 4 = (2u + 8a) - (2u + 5a) \\\ \Rightarrow 1 = 2u + 8a - 2u - 5a \\\ \Rightarrow 1 = 3a \\\ \Rightarrow a = \dfrac{1}{3}m/{\sec ^2} \\\$
This is the acceleration produced in the particle.
Putting this value in equation (i),
$5 = 2u + 8a \\\ \Rightarrow 5 = 2u + 8 \times \dfrac{1}{3} \\\ \Rightarrow 5 - \dfrac{8}{3} = 2u \\\ \Rightarrow 2u = \dfrac{{15 - 8}}{3} \\\ \Rightarrow 2u = \dfrac{7}{3} \\\ \Rightarrow u = \dfrac{7}{6}m/\sec \\\$
It is the initial velocity of a particle.
Now, we are finding distance travels in the next $2\sec$. So, total time will be $8 + 2 = 10\sec$
So, the second equation of motion $S = ut + \dfrac{1}{2}a{t^2}$
Putting $u = \dfrac{7}{6}m/\sec$, $a = \dfrac{1}{3}m/{\sec ^2}$and $t = 10\sec$
$s = \dfrac{7}{6} \times 10 + \dfrac{1}{2} \times \dfrac{1}{3} \times {10^2} \\\ \Rightarrow s = \dfrac{{70}}{6} + \dfrac{{100}}{6} \\\ \Rightarrow s = \dfrac{{170}}{6} \\\ \Rightarrow s = 28.33m \\\$
It is the total distance travelled by body in $10\sec$.So, the displacement of body in last $2\sec$.

$${s_2} = {s_{10}} - {s_8} \\\ \Rightarrow {s_2} = 28.33 - 20 \\\ \Rightarrow {s_2} = 8.33m \\\$$

Hence, the distance in next $2\sec$ is $8.33m$.

Note: - Here, we use the second equation of motion. If there is a varying acceleration, then we can use the third equation of motion. One thing is to be remembered then when the velocity is increasing then there is acceleration and if velocity is decreasing then this acceleration is known as retardation and taken as negative.