Question

A drag racing car starts from rest at t=0 and moves along a straigh line with velocity given by v=bt^2 , where b is constant the expression for the distance traveled by this car from its position at t=0

Answer 1

The distance travelled can be as x = bt³/3. Just integrate the expression of velocity
since v = dx/dt And the distance travelled at t=0 is x= b(0)³/3 = 0

Answer 2

Heya!!Here is your answer friend ⤵⤵

Given that , u=0. ( as it starts from rest )
t=0
v=
$bt {}^{2}$

Now we shall differentiate the given velocity ,

Using the rule of differentiation,


$x {}^{n} = nx {}^{n - 1}$
We have , v= 2t

Distance at t=0

v=0

Distance travelled is denoted by S

Using second equation of motion ,

$s = ut + \frac{1}{2} at {}^{2}$

We have

S= 0


Hope it helps you ✌✌