A certain automobile can decelerate at |a1| = 1.6 m/s2. Traveling at a constant

Question

A certain automobile can decelerate at |a1| = 1.6 m/s2. Traveling at a constant v1 = 29 m/s, this car comes up behind a car traveling at a constant v2 = 4 m/s.

a) What is the speed of the faster car relative to the slower car?

b) How close to the slower car can the driver of the faster car come before applying his brakes and still avoid a collision?

c) At what time does the inevitable collision of the two cars occur?

d) How far beyond its position at t = 0 does the slower car get before it is hit?

Explanation / Answer

a) The relative speed of faster car relative to smaller car=29-4=25m/s

b)Let x be the minimum distance to avoid collision.

We can say that the collision will be avoided when be reach relative speed=0.

relative distance travelled in the process =-u2/-2a=252/(2*1.6)=195.3 m

c)the collision occurs when the relative distance covered=30m

30=25t-0.5*1.6t2

t=1.25 s (a higher value of t is not applicable since the collision already occurs at this point)

d)distance travelled by the slower car before being hit=1.25*4=5m

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