Explanation:
It is given by :
\(s=\dfrac{1}{2}at^2\)..... (1)
Where
s = distance covered
a = acceleration of an object
t = time taken
We need to manipulate given equation to find the value of t. On cross multiplication to both side of equation (1) we get:
\(2s=at^2\)
Dividing both side of above equation by a we get :
\(t^2=\dfrac{2s}{a}\)
Taking square root of above equation we get :
\(t=\sqrt{\dfrac{2s}{a}}\)
So, the value of t is, \(t=\sqrt{\dfrac{2s}{a}}\). Hence, this is the required solution.