Question

A body is travelling with uniform acceleration and travels $84\,m$ in first $6\sec$ and $180\,m$ in the next$5\sec$. Find the velocity and acceleration of the body.

Answer 1

Here, we have been given the information about a body which is traveling with uniform acceleration but the two velocities at different time durations are given for that respective body. So by using this information we have to discuss the velocity and acceleration of that body.

Formula used:
$s = ut + \dfrac{1}{2}a{t^2}$
$v = u + at$
Where, $s$ is the distance traveled by the body, $u$ is the initial velocity of the body, $v$ is the final velocity, $a$ is the acceleration of the body and $t$ is the time taken by that body to cover certain distance.

Complete step by step answer:
Let us first consider the given information as: let $u$ be the initial velocity of the body, $v$ be the final velocity of the body at the end of $6\sec$ and $v'$ be the final velocity of the body and $a$ be the acceleration of the body. For the body at time $t = 6\sec$. The distance formula for the body that is travelled in time $t = 6\sec$ is given by:
$s = ut + \dfrac{1}{2}a{t^2}$...........….. (Kinematic equation)
$\Rightarrow {s_1} = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow 84 = u(6) + \dfrac{1}{2}a{(6)^2}$
$\Rightarrow 84 = 6u + 18a$.........…... $(1)$
Final velocity of the particle at $t = 6\sec$
$v = u + 6a$
After using this relation, after $6\sec$ in the second part of the motion, the required equation is given by considering this final velocity in the first part as initial velocity of the second part.
${s_2} = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow 180 = \left( {u + 6a} \right)5 + \dfrac{1}{2}a{(5)^2}$......…... (Consider $u$ as $v$ here.)
$\Rightarrow 180 = \left( {u + 6a} \right)5 + \dfrac{1}{2}a(25)$
On simplification of this equation we get:
$180 = 5u + \dfrac{{85}}{2}a$............…. $(2)$
Now we have to solve equation $(1)$ and $(2)$ such that we obtain:
From equation$(1)$, we have:
$u = 14 - 3a$...........…. $(3)$
By using this value in equation$(2)$, we get
$180 = 5(14 - 3a) + \dfrac{{85}}{2}a$
$\Rightarrow 180 = 70 - 15a + \dfrac{{85}}{2}a$
$\Rightarrow 180 - 70 = \dfrac{{ - 30a + 85a}}{2}$
$\Rightarrow 110 = \dfrac{{55a}}{2}$
$\Rightarrow a = 4\,m{s^{ - 2}}$
By using this value in equation$(3)$, we get
$u = 14 - 3(4) = 2\,m{s^{ - 1}}$
Therefore, we obtained the acceleration and initial velocity of a body such that:
$\boxed{a = 4\,m{s^{ - 2}}}$ and $\boxed{u = 2\,m{s^{ - 1}}}$

Note: Here, we see that by using kinematic equation we can be able to calculate the unknown values in the problem such that if any of the two or three value in the equation are known to us it would be easier to find the remaining term but here we have seen the different scenario that there are two unknowns in the each of the equations and we have to calculate both of them. For this we have calculated both the conditions and put them into suitable equations and on simplifying we obtained the answer. Remember that each question is different, you must remember the concept of kinematics very well.