A car travels at 108 kph from A

Answered question

2022-03-26

A car travels at 108 kph from A to B for t seconds, applies the brakes for 4 seconds between B and C to give the car a constant deceleration and a speed of 72 kph at C, and finally travels at this speed from C to D, also for t seconds. If the total horizontal distance from A to D is 3100 m, determine the distance between A and B.

Answer & Explanation

nick1337

nick1337

Expert2022-07-03Added 777 answers

Speed of car between A and BVAB=108kmh=30ms time taken by the car to travel from A to B = t distance travalled by the car between A and B is SAB=VABtSAB=30t

Breakes applies to t=4s so that the speed of car at C becomes 72 kmph Vc=72kph=20ms

Speed at B is Vb=30ms acceleration of the car between B&C in a=VC-VBt

a=20-504a=-2.5ms2

distance travelled by the car between B and C in

SBC=VC2-VB22R=202-3022×((-2)×5)=-500-5 SBC=100m

speed of the car between C and D is VCD=20ms

distance travelled by the car between C and D is SCD=VCPtSCD=20t

Total distance, SAB=3100m

but, SAB=SAB+SBC+SCD=30t+100+20t

3100=50t+100  50t=3000   t=310050=60$

distance, SAB=30t=30×60=1800m

distance travelled by the car between A and B is SAB=1800m

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