Question

If velocity v of a particle moving on a straight line as a function of time t is given as v=5-tm/s then the distance covered by the particle is

Answer 1

SUPPOSE THE QUESTION IS ASKING YOU THE DISTANCE TRAVELLED AFTER 2 SECONDS (just suppose ) . THEN THERE ARE TWO WAYS YOU CAN SOLVE IT ; FIRST ONE IS ,

since velocity is given as a function of time as 5t m/s ( integration of velocity ) 5t m/s

the integrated diplacement would be 5t^2/2 .

now put the value of t . that would be your displacement . IN THE EXAMPLE WE TALKED ABOUT TIME =2s. SOLVING FOR THAT , YOU WOULD GET (5)(2^2)/2 WHICH GIVES 10m . Similarly you can attempt these questions .

Answer 2

Answer:

The distance covered by the particle is $x=5t-\dfrac{t^2}{2}+C$

Explanation:

Given that,

Velocity v = 5-t m/s

We need to calculate the distance covered by the particle

The velocity is the first derivative of the position of the particle.

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

When we integrate to the velocity then we found the distance of the particle.

$\int_{0}^{x}{dx}=\int_{0}^{t}({5-t})dt$

On integration w.r.to t

$x=5t-\dfrac{t^2}{2}+C$

Hence, The distance covered by the particle is $x=5t-\dfrac{t^2}{2}+C$