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3 years ago

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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?

2 answers:

dolphi86 [110]3 years ago

6 0

Answer:

t = 11.83 s

It will take 11.83 seconds

Step-by-step 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

notsponge [240]3 years ago

4 0

Answer:

It takes 11.83 s.

Step-by-step 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

$t=\frac{distance}{relative \:speed}$

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 $115 \:m\cdot (\frac{1\:km}{1000\:m} )=0.115\:km$ .

The relative speed is

$55.0\:\frac{km}{h} -20.0\:\frac{km}{h}=35\:\frac{km}{h}$

and the time when the two cars meet is

$\frac{0.115 \:km}{35\:\frac{km}{h} } \approx0.003286 \:h$ or $0.003286 \:h\cdot (\frac{3600}{1\:h} )=11.83 \:s$

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C and E

Step-by-step explanation:

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