Question

(a) Explain the meaning of the following equation of motion: $v = u + at$, where symbols have their usual meanings.
(b) A body starting from rest travels with uniform acceleration. If it travels 100m in 5s, what is the value of acceleration?

Answer 1

In order to denote the position of an object we define a coordinate system and an origin as a reference point. If a body moves from one position to another position then subtracting the initial position vector from the final position vector gives us displacement. Rate of change of displacement with respect to time gives us velocity.

Formula used:
\eqalign{ & \dfrac{{dv}}{{dt}} = a \cr & \int {(a)} dt = v \cr}
$v = u + at$
$s = ut + \dfrac{1}{2}a{t^2}$

Complete answer: or Complete step by step answer:
Rate of change of displacement with respect to time gives us the velocity and the rate of change of velocity with respect to time gives us the acceleration. Now if we are given with acceleration as a function of time and we were asked to find out the displacement then first we should get the velocity function by integrating the acceleration with in the proper limits given and then later after getting the velocity function, integrate it again with respect to time to get the displacement.
\eqalign{ & \dfrac{{dv}}{{dt}} = a \cr & \int {(a)} dt = v \cr}
Where ‘v’ is the velocity and ‘a’ is the acceleration and ‘t’ is the time.
$\int {(a)} dt = v$
\eqalign{ & \Rightarrow \left[ v \right]_u^v = \left[ {at} \right]_{t = 0}^{t = t} \cr & \Rightarrow v - u = at \cr & \Rightarrow v = u + at \cr}
So in the above equation we have ‘v’ is the velocity at time instant ‘t’ and ‘a’ is the uniform acceleration and ‘u’ is the initial velocity.
For solving part b we have the equation
$s = ut + \dfrac{1}{2}a{t^2}$
Where ‘s’ is the displacement which is given as 100m and u=0 and t=5s. Hence by substituting all these values we get
$s = ut + \dfrac{1}{2}a{t^2}$
\eqalign{ & \Rightarrow 100 = 0 + \dfrac{1}{2}a{(5)^2} \cr & \Rightarrow \dfrac{{200}}{{25}} = a \cr & \Rightarrow a = 8m/{s^2} \cr}
So acceleration of body is $8m/{s^2}$

Note:
The second formula which we had used is only valid if the acceleration is uniform. If it is not uniform then that will be a function of time and we have to integrate acceleration too with respect to time and we get entirely different results. Both the formulas are valid for uniform acceleration only.