Question

If a particle has velocity v=t^2-5t+6 then find average velocity in 5sec​

Answer 1

Answer:

6 . 5 m

Explanation:

The relation is = v where x is displacement dt and v is velocity which is given as v = 3 - t

so we have dx = vdt

after 3sec the direction of motion will get reversed.

SO

on integrating both sides we get fx dx = (3 t)dt = 9 9 2 = 4.5meter

now integrating both sides for t = 3 to t = 5 we have the displacement as

SX45 dx = 5(3 t)dt = − 2meter -4.5 so the net distance will be X = (2) + 4.5 =

6.5meter

Answer 2

x=2t
2
−5t+6
⇒v=
dt
dx
​
=4t−5
⇒v=7ms
−1
att=3seconds.
Now,Positionint=3seconds
⇒x=18−15+6=9meters.
and position at t=0 seconds is 6m⇒Averagevelocity=
3
9−6
​
=1ms
−1