Solveeit Logo

Question

Question: If a body travels \(30\,m\) in an interval of \(2s\) and \(50\,m\) in the next interval of \(2s\) th...

If a body travels 30m30\,m in an interval of 2s2s and 50m50\,m in the next interval of 2s2s then the acceleration of the body is
A. 10ms210\,m{s^{ - 2}}
B. 5ms25\,m{s^{ - 2}}
C. 20ms220\,m{s^{ - 2}}
D. 25ms225\,m{s^{ - 2}}

Explanation

Solution

In physics, the rate of change of distance covered by a body with respect to time is known as velocity and thus the rate of change of velocity with respect to time is known as acceleration of the body, here, we will use the newton’s equation of motion to determine the magnitude of acceleration of the body.

Formula used:
According to newton’s equation of motion we have,
S=ut+12at2S = ut + \dfrac{1}{2}a{t^2}
Where S,u,a,tS,u,a,t are the distance covered by the body, initial velocity of the body, acceleration of the body, time interval of the body.

Complete step by step answer:
Let us suppose the initial velocity of the body is uu its acceleration aa and according to the question for time period t=2st = 2s the distance covered by the body is S=30mS = 30\,m.
Using the formula, S=ut+12at2S = ut + \dfrac{1}{2}a{t^2} we have,
30=2u+12×4a30 = 2u + \dfrac{1}{2} \times 4a
30=2u+2a\Rightarrow 30 = 2u + 2a
15=u+a(i)\Rightarrow 15 = u + a \to (i)
According to the question, in next 2s2s which means total time t=2+2=4st = 2 + 2 = 4s body covers extra 50m50m which means total distance covered from initial point will be S=50+30=80mS = 50 + 30 = 80\,m so using again the formula, S=ut+12at2S = ut + \dfrac{1}{2}a{t^2} we have,
80=4u+8a80 = 4u + 8a
20=u+2a(ii)\Rightarrow 20 = u + 2a \to (ii)
From equations i(and)iii(and)ii subtract equation i(from)iii(from)ii we get,
2015=2aa20 - 15 = 2a - a
a=5ms2\therefore a = 5\,m{s^{ - 2}}
So, the acceleration of the body will be of magnitude 5ms25\,m{s^{ - 2}}.

Hence, the correct option is B.

Note: It must be remembered that, while solving such question always take total distance for the value of S in the formula S=ut+12at2S = ut + \dfrac{1}{2}a{t^2} from initial point to the time when its mentioned in the question, Acceleration is a vector quantity and its mathematically simply written as a=dvdt\vec a = \dfrac{{d\vec v}}{{dt}} and also when body has a negative acceleration which means body has a decreasing acceleration its known as deceleration.