Attempt 1 of Unlim Points Available: 1 Assignment Policy Spi Time Spent 00:00:35
ID: 1657339 • Letter: A
Question
Explanation / Answer
Given
speed of car A v1 = 29.9 km/h = 8.305 m/s (x kmph*5/18 = y m/s)
speed of car B v2 = 31.8 km/h = 8.833 m/s
car A moving in the -ve direction
car B moving in the +ve direction
a) the dog is at rest and observing the cars
velocity of car b relative to dog is V2 = 8.833 m/s
b) as the cars are moving in the opposite directions then the relative velocity of the cars is
V = V1+V2 = 8.305+8.833 m/s = 17.138 m/s
c) initially the cars are separted by 0.12 km = 120 m
time taken by them if the separation between them is 1.04 km = 1040 m
so the distance they should travel furthur is 1040-120 m = 920 m
let time taken by car A to cover a half of the distance is t1 = 460/8.305 s = 55.388 s
time taken by the car b to cover remainign 460 m is t2 = 460/8.833 s = 52.077 s
so total time T is = 55.388 + 52.077 s = 107.465 s
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