Question

If velocityv of a particle moving on a straight line as a function of time is given as v=5-t(m/s) then the distance covered by the particles in first 10s​

Answer 1

Answer:

Given:

Velocity varies with time according to the following relationship :

$\boxed{ \huge{ \red{ \sf{v = 5 - t}}}}$

To find:

Distance covered by the particle in 10 seconds.

Concept:

In this type of functions , we need to find distance by performing definite integration by putting the limits.

$\boxed{ \huge{ \red{ \int dx = \int v \: dt}}}$

Calculation:

$\therefore \displaystyle \int \: dx = \int v \: dt$

Putting the limits :

$= > \displaystyle \int_{0}^{x} \: dx = \int_{0}^{10} v \: dt$

$= > \displaystyle \int_{0}^{x} \: dx = \int_{0}^{10} (5 - t) \: dt$

$= > x = \bigg \{5t \bigg \}_{0}^{10} - \bigg \{ \dfrac{ {t}^{2} }{2} \bigg \}_{0}^{10}$

$= > x = 0 \: m$

So displacement is zero.

For calculating the distance , draw the graph first .

Total distance be d :

$d = area \: of \: 2 \: triangle$

$= > d = (\frac{1}{2} \times 5 \times 5) \times 2$

$= > d = 25 \: metres$