Question

A uniformly accelerating body covers a distance of $10{\text{ m}}$ in the $3{\text{ rd}}$ second and $16{\text{ m}}$ in the $6{\text{ th}}$ second. What is the initial velocity and acceleration?

Answer 1

Use the formula to determine the distance of the particle in nth second as determine the initial velocity and acceleration of the particle. Use the kinematic relation to determine the distance travelled by the particle in the next 3 second using quantities you obtained above.

Formula used:
${s_n} = u + \dfrac{1}{2}a\left( {2n - 1} \right)$
Here, $u$ is the initial velocity of the body, $a$ is the uniform acceleration and $n$ is the value of the second for which the distance is to be calculated.

Complete step by step answer:
We know that the displacement of the uniformly accelerated body is given by the relation,
$\Rightarrow {s_n} = u + \dfrac{1}{2}a\left( {2n - 1} \right)$
Here, u is the initial velocity of the body and a is the uniform acceleration.
We have given, the distance is $10{\text{ m}}$in the $3{\text{ rd}}$ second. Therefore, we can write
$\Rightarrow 10 = u + \dfrac{1}{2}a\left( {2(3) - 1} \right)$
$\Rightarrow 10 = u + \dfrac{5}{2}a - - - - - - - - - - (1)$

Also, the distance is $16{\text{ m}}$ in the $6{\text{ th}}$ second. Therefore, we can write,
$\Rightarrow 16 = u + \dfrac{1}{2}a\left( {2(6) - 1} \right)$
$\Rightarrow 16 = u + \dfrac{{11}}{2}a - - - - - - - - - (2)$
Subtract equation (1) from equation (2)
$\Rightarrow 16 - 10 = u + \dfrac{{11}}{2}a - \left( {u + \dfrac{5}{2}a} \right)$
$\Rightarrow 6 = 3a$
$\therefore a = 2{\text{ m/}}{{\text{s}}^2}$
Solve the simultaneous equations (1) and (2) for $u$, we get
$\therefore u = 5{\text{ m/s}}$

Hence,the initial velocity and acceleration are $5{\text{ m/s}}$ and $2{\text{ m/}}{{\text{s}}^2}$.

Note: Since the acceleration is uniform, the acceleration is the same at any position. The final velocity at a given second is also the initial velocity of the particle at that second due to uniform acceleration. The formula for determining the distance of the particle at nth second is obtained from the relation $s = ut + \dfrac{1}{2}g{t^2}$.