Question

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4 and 5 both

Answer 1

Answer:

1.We know that s = ut + 1/2at²

According to question, u = 0 [As it was in rest]; a = 3ms⁻² ; t = 8s

s = 0 x 8 + 1/2 x 3 x 8 x 8

s = 3 x 4 x 8

s = 96m

2.Given : 

First Car  A: 

Initial Velocity= u= 52 km/hr=52x5/18=14.4 m/s

Final velocity= V= 0m/s

time =t=5sec

Refer attachment for Graph :

Distance travelled : Area of traingle AOB =1/2 OB x AO

=(1/2)x 14.4 x 5=72/2=36m

Second Car  B:

Initial Velocity= u= 3 km/hr=3x5/18=0.83m/s

Final velocity= V= 0m/s

time =t=10sec

Refer attachment for Graph :

Distance travelled : Area of traingle COD =1/2 OD x CO=(1/2)x 0.83 x 10=4.1m

So, after applying of brakes , Car A has travelled a distance of 3m and car B has travelled a distance of 4.1m.

∴ Car B has travelled farther than Car A after the brakes are applied.

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Answer 2

Answer: Here is the answer for 4th and 5th question