Question

A particle moves in a straight line. The displacement x of the particle varies with time as x = 2 – 5t + 6t2. Then the initial velocity of the particle is

Answer 1

The relation between displacement and velocity is given as:

$v = \frac{dx}{dt}$

Here we are given displacement as a function of time. 


$x = 2+5t+6t^2 \\ \\ \implies x = 6t^2+5t+2$


We can differentiate it with respect to time to get velocity.

$v = \frac{dx}{dt} \\ \\ \\ \implies v = \frac{d}{dt} (6t^2+5t+2) \\ \\ \\ \implies v = 12t+5$


Now, we have to find initial velocity. Whenever we are asked to find initial conditions, we have to take t = 0.

So, we can find initial velocity by putting t = 0

$v = 12t+5 \\ \\ \implies v = 12(0)+5 \\ \\ \implies v = 0 + 5 \\ \\ \\ \implies \boxed{\bold{v = 5 \, \, m/s}}$


Thus, The Initial Velocity is 5 m/s.

Answer 2

Considering the equation is x = 2 +5t +6t²
x= 6t² + 5t + 2
We know that,
v=dx / dt

So,
v= d(6t² +5t +2) /dt
v= 12t + 5
For, initial velocity substitute t=0;
v= 12(0) +5
v= 5m/sec

So, the initial velocity was 5m/sec