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The driver of a car travelling at a speed of 20m/sec, wishes to overtake a truck that is moving with a constant speed of 20 m/sec in the same lane. The cars maximum acceleration is 0.5 m/s^2 . Initially the vehicles are seperated by 40m, and the car returns back into its lane after it is 40 m ahead of truck. The car is 3 m long and the truck 17m. (a) Find the minimun time required for the car to pass the truck and return back to its lane? (b) What distance does the car travel during the time? (c) What is the final speed of the car?

The driver of a car travelling at a speed of 20m/sec, wishes to overtake a truck that is moving with a constant speed of 20 m/sec in the same lane. The cars maximum acceleration is 0.5 m/s^2 . Initially the vehicles are seperated by 40m, and the car returns back into its lane after it is 40 m ahead of truck. The car is 3 m long and the truck 17m.


(a) Find the minimun time required for the car to pass the truck and return back to its lane?


(b) What distance does the car travel during the time?


(c) What is the final speed of the car?

Grade:12

2 Answers

neeraj agarwal
34 Points
9 years ago

i)for min time the car should use max accln and now car will travel 40+40+3+17=100m                                                                    using newtons 2nd eqn                                                                                                                                                                                     100=v(rel)t=1/2at2                                                                                                                                                   100=0+t2/4                                                                                                                                                                                          t=20s                                                                                                                             ii)100m                                                                                                                                                                                           using newtons 1st eqn                                                                                                                                             v=20+0.5*20=30m/s

RAJEEV CHOUDHARY
9 Points
9 years ago

a) For minimun time we can take maximum acceleration and also that car will be in accelerated motion throughout its overtake.

Distance travelled by car during overtake S1 = 20t +0.5*0.5 t2

Distance travelled by truck S2 = 20t

S1-S2 = 80

So t2 = 320

b)Substituting the value of t in above equation for S1

c) V = 20 + 0.5t.

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