Question

Velocity of a moving particle varies with time given by v=t^2 +2t +2 . If position of the particle at t=0 is 2m then position of the particle at t=2 sec will be

Answer 1

Position of particle at t=2 seconds will be 17 m.

Explanation:

You've to just put the value of t=2 in above question.

And it will be the velocity of body at t=2.

But we have to find distance travelled by it in 2 seconds.

And Velocity at t=0,

t² + 2t + 2

0² + 2×0 + 2

0 + 0 + 2

2 m/s

Distance Covered For 0th Second = 2m

So, velocity at t=1,

t² + 2t + 2

1² + 2×1 + 2

1 + 2 + 2

5 m/s.

Distance Covered For 1th Second = 5m

So, velocity at t=2,

t² + 2t + 2

2² + 2×2 + 2

4 + 4 + 2

10 m/s.

Distance Covered For 0th Second = 10m

Total distance Covered by end of 2 seconds will be 2+5+10 = 17 m

which means, position of particle at t=2 seconds will be 17 m.

Hope It Helps ;)

Answer 2

Given :-

Velocity of particle as a function of time.

v = t² + 2t + 2

Time, at t = 0 is 2m distance travelled.

Now we are required to find the distance travelled at time, t = 2s. Here we will integrate the function with respect to time.

$\int{dS}\:= \int{v\:dt}$

${S}\:= \int{{t}^{2}+2t+2\:dt}$

${S}\:= \dfrac{{t}^{3}}{3} + {t}^{2} + 2t$

Putting, t = 2 s.....

$\bf{S = \dfrac{{t}^{3} + 3{t}^{2} + 6t}{3}}$

$\bf{S = \dfrac{32}{3}m}$

Hence,

The distance travelled = S = 32/3 m