Question

The velocity of a car decreases from 30m/s to 15m/s,when it travels a distance of 100m.The distance travelled from this position before it comes to rest is ____. Please help me with this answer...

Answer 1

Answer:

First, you have to work out the acceleration. We know the initial velocity (30 m/s) and the final velocity (15 m/s), and the distance over which the deceleration happened (75 m), so we can work out the acceleration:

v2f=v2i+2aΔx

(15m/s)2=(30m/s)2+2a(75m)

225m²/s²=900m²/s²+(150m)a

−675m²/s²=(150m)a

−4.5m/s²=a

Now that we know the acceleration, we can use the same equation to figure out the displacement when vf equals 0 m/s:

(0m/s)2=(30m/s)2+2(−4.5m/s²)Δx

0m²/s²=(900m²/s²)+(−9m/s²)Δx

−900m²/s²=(−9m/s²)Δx

100m=Δx

So there you have it. The particle will come to a stop after traveling 100 meters.

OK BRO.....

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Answer 2

Answer:

200m

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