Question

A 1000-kg car moves as speed from 10 m/s to 30 m/s in 20 seconds. What is the average speed? What is the acceleration? What is the average force acting upon it? If the car turns along a circle at speed 5 m/s around a curve 30 meters in radius, how big the centripetal force is needed?

Answer 1

◆【Solution】◆

Let we make you clear that the average speed is independent of the no. of speeds like we can not calculate the average like sum of magnitude divide by no. of magnitude.

Average speed = ∆v/∆t

where ∆v is the magnitude of velocity i.e speed ∆t is the time taken to change speed by ∆v.

So initial velocity (u) = 10m/sec

Final velocity (v) = 30m/sec

∆v = v - u

∆v = (30 - 10)m/sec

◆【∆v = 20m/sec】◆

∆t = 20sec

◆【Average speed = 20/20 = 1m/sec】◆

Acc to Newton

Net force acting on a body is equal to rate of change of linear momentum

●【Force = ∆p/∆t】●

∆p = final momentum - Initial momentum

∆p = (mv - mu)

∆t = 20sec

force = m(v - u)/20

Force = (1000)(1)

◆【Force = 1000N】◆

Centripetal force = mv²/r

Centripetal force = (1000)(5)²/30

Centripetal force = 2500/3

Answer 2

Thus the magnitude of centripetal force is 833 N.

Explanation:

Average speed = ∆v/∆t

• Where ∆v is the magnitude of velocity i.e speed
• ∆t is the time taken to change speed by ∆v.

Initial velocity (u) = 10 m/s

Final velocity (v) = 30 m/s

∆v = v - u

∆v = (30 - 10)m/sec

∆v = 20 m/s

∆t = 20 s

Average speed = 20/20 = 1 m/s

According to Newton law of motion:

Net force acting on a body is equal to rate of change of linear momentum

Force = ∆p/∆t

∆p = final momentum - Initial momentum

∆p = (mv - mu)

∆t = 20 s

Force = m(v - u)/20

Force = (1000)(1)

Force = 1000 N

Centripetal force = mv²/r

Centripetal force = (1000)(5)²/30

Fc = 2500 / 3 = 833 N

Thus the magnitude of centripetal force is 833 N.