Two 4 meter lorry cars are moving along the same direction on a straight road with a constant speeds of 70 km/hr and 90km/hr. if a passenge falls asleep when the faster car is 20m behind the slower car but wakes up after it is 20m ahead of the slower car. find how long he slept?
Multiply each speed by .2777 to get them in m/s. You'll car 1 is gaining 5.6 meters each second. For the 40m you're talking about a 7 second nap.
I need the calculation
time = 8.6secs
To solve this problem, we need to analyze the relative motion of the two cars and determine the time it takes for the faster car to travel a distance of 40 meters (from being 20 meters behind the slower car to being 20 meters ahead).
First, let's convert the speeds of the cars from km/hr to m/s:
Speed of the slower car = 70 km/hr = (70 * 1000) / (60 * 60) = 19.44 m/s
Speed of the faster car = 90 km/hr = (90 * 1000) / (60 * 60) = 25 m/s
Since the cars are moving in the same direction, their relative velocity is the difference between their speeds:
Relative velocity = Speed of the faster car - Speed of the slower car = 25 m/s - 19.44 m/s = 5.56 m/s
Now, let's calculate the time it takes for the faster car to travel a distance of 40 meters with a velocity of 5.56 m/s:
Time = Distance / Velocity = 40 meters / 5.56 m/s = 7.19 seconds
Therefore, the passenger slept for approximately 7.19 seconds.