Question

A body starts with some initial velocity and a constant acceleration. It covers a distance of $200m$ in the first four seconds and a distance of $220m$ in the next two seconds. The acceleration of the body and its velocity at the end of seventh second are
(A) $a = 5m{s^{ - 2}},v = 75m{s^{ - 1}}$
(B) $a = 7.5m{s^{ - 2}},v = 100m{s^{ - 1}}$
(C) $a = 20m{s^{ - 2}},v = 150m{s^{ - 1}}$
(D) $a = 150m{s^{ - 2}},v = 10m{s^{ - 1}}$

Answer 1

Hint:In order to solve this question, we will use the equation of motion to relate the distance covered by the body in first four second and then form another equation relating distance covered in next two second and on solving the two formed equation, we will solve for acceleration and velocity of the body at end of seventh second.



Formula used:
Two of the equation of motion are
$v = u + at$ where u is initial velocity and v is final velocity during time ‘t’ with acceleration ‘a’ and
$S = ut + \dfrac{1}{2}a{t^2}$ where,
S denotes distance covered by the body during time ‘t’ with initial velocity ‘u’ and acceleration of the body is ‘a’.


Complete answer:
According to the question, we have given that a body covers a distance of $200m$ in first four second which is $t = 4s$, and let initial velocity of the body is u and constant acceleration is ‘a’ so using the formula $S = ut + \dfrac{1}{2}a{t^2}$ we get,
$
  200 = u(4) + \dfrac{1}{2}(a){(4)^2} \\
  200 = 4u + 8a \to (i) \\
 $
Now, it’s give that in next two seconds body covers distance of $220m$ so, in total period of time $t = 4 + 2 = 6s$ body covers a total distance of $S' = 200 + 220 = 420m$ so again using the formula $S = ut + \dfrac{1}{2}a{t^2}$ we get,
$
  420 = u(6) + \dfrac{1}{2}(a){(6)^2} \\
  420 = 6u + 18a \to (ii) \\
 $
On solving equations (i) and (ii) by substituting method we get,
$
  420 = 6(\dfrac{{200 - 8a}}{4}) + 18a \\
  840 = 600 - 24a + 36a \\
  240 = 12a \\
   \Rightarrow a = 20m{s^{ - 2}} \\
 $
and so, $u = 10m{s^{ - 1}}$
So, acceleration of the body is $a = 20m{s^{ - 2}}$
and velocity at the end of seventh second $t = 7$ and be calculated using the formula $v = u + at$ we get,
$
  v = 10 + (20)(7) \\
  v = 150m{s^{ - 1}} \\
 $

Hence, the correct option is Option (C).


Note:While solving such question never makes a mistake taking given distance in next ${t^{th}}$ second as a total distance in time ‘t’ always add the total distance to the time ‘t’ while putting value in newton’s equation of motion and solve carefully such numerical equations to avoid any minor calculation mistakes.