Question

How long does it take an automobile traveling in the left lane of a highway at 55.0 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 115 m apart?

Answer 1

It takes 11.83 s.

Explanation:

If two objects move in the same direction at different speeds their relative speed = Fastest speed – slowest speed.

The time after which the two objects meet is given by

`t=(distance)/(relative \:speed)`

We know that an car traveling in the left lane of a highway has a speed of 55.0 km/h and the other car has a speed of 20.0 km/h.

The initial distance between them was 115 m, which is equivalent to
`115 \:m\cdot ((1\:km)/(1000\:m) )=0.115\:km` .

The relative speed is

`55.0\:(km)/(h) -20.0\:(km)/(h)=35\:(km)/(h)`

and the time when the two cars meet is

`(0.115 \:km)/(35\:(km)/(h) ) \approx0.003286 \:h` or
`0.003286 \:h\cdot ((3600)/(1\:h) )=11.83 \:s`

Answer 2

t = 11.83 s

It will take 11.83 seconds

Explanation:

The distance covered can be derived using the equation of motion with no acceleration.

distance d = speed v × time t

d = vt

For the first automobile,

v = 55km/h

d1 = 55t ....1

For the second automobile,

v = 20km/h

Initial distance d0 = 115m = 0.115km

d = vt + d0

d2 = 20t + 0.115 ......2

For them to meet/overtake each other,

Their distance at that time must be equal.

d1 = d2

55t = 20t + 0.115

55t-20t = 0.115

35t = 0.115

t = 0.115/35 hr

Converting to seconds

t = 0.115/35 × 3600s

t = 11.83 s