How long does it take an automobile traveling in the left lane of a highway at 70.0 km/h to overtake (become even with) another car that is traveling in the right lane at 20.0 km/h when the cars' front bumpers are initially 110 m apart?
s
let the distance needed to pass be x km
so the time taken by the slower car = x/20
and the time taken by the faster car = (x+.110)/70
but those times are equal, so ....
x/20 = (x+.110)/70
70x = 20x + 2.2
50x = 2.2
x = 2.2/50 = .044
time taken by slower car to be overtaken = .044/20
= .0022 hours
= 7.92 seconds
notice .044/20 = .0022
and (.044 + .110)/70 = .0022
To determine how long it takes for the automobile in the left lane to overtake the car in the right lane, we can use the relative velocity between the two cars.
The relative velocity is calculated by subtracting the velocity of the car in the right lane from the velocity of the car in the left lane. In this case, the relative velocity is (70.0 km/h - 20.0 km/h) = 50.0 km/h.
First, we need to convert the relative velocity from kilometers per hour to meters per second. To do this, we divide the relative velocity by 3.6 (since there are 3.6 seconds in an hour).
Relative velocity = 50.0 km/h รท 3.6 = 13.89 m/s.
Now that we have the relative velocity, we can calculate the time it takes for the cars to become even with each other. To do this, we use the formula:
Time = Distance / Relative velocity.
The distance between the cars' front bumpers is given as 110 m. Plugging in the values:
Time = 110 m / 13.89 m/s = 7.92 seconds.
Therefore, it takes approximately 7.92 seconds for the automobile in the left lane traveling at 70.0 km/h to overtake the car in the right lane traveling at 20.0 km/h, assuming the cars' front bumpers are initially 110 m apart.