A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.5 m/s 2. A green car arrives at the position of the stop-light 5 s after the light had turned green. If t = 0 when the light turns green, at what time does the green car catch the blue car if the green car maintains the slowest constant speed necessary to catch up to t

Question

Physics

Bishop

18 February, 23:27

A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.5 m/s 2. A green car arrives at the position of the stop-light 5 s after the light had turned green. If t = 0 when the light turns green, at what time does the green car catch the blue car if the green car maintains the slowest constant speed necessary to catch up to t

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Amani Mathis · Answer 1

The green car will catch the blue one after 10 s of the stop-light turning green.

Explanation:

The equation for the position of objects moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t

x0 = initial position

t = time

a = acceleration

v0 = initial velocity

For the blue car, this will be its position at time t

x = 0 m + 0 m/s · t + 1/2 · 0.5 m/s² · t²

x = 0.25 m/s² · t²

For the green car, that moves at constant speed, its position will be:

x = 0 m + v · t (where v = velocity)

x = v · t

Now let's find how much distance the blue car has traveled until the green car arrives at the stop-light:

x = 0.25 m/s² · (5s) ² = 6.25 m

When the persecution begins (5 s after the stop-light turns green), the blue car is 6.25 ahead of the green car.

When the green car catches the blue one, its position is the same as the position of the blue car.

Then:

Position of the blue car = position of the green car

6.25 m + 0.25 m/s² · t² = v · t

0.25 m/s² · t² - v · t + 6.25 m = 0

Let's use the formula for solving quadratic equations:

a = 0.25

b = - v

c = 6.25

(-b ± √ (b²-4·a·c)) / 2·a

Let's replace with the dа ta:

(v ± √ (v² - 4 · 0.25 m/s² · 6.25 m)) / 2· 6.25 m

(v ± √ (v² - 6.25 m²/s²)) / 12.5 m

The minimum value of v that solves this equation is the value that makes the content inside the root to be null. A lower value than that and the content inside the root will be negative and have no solution (real solution).

Then:

v² - 6.25 m²/s² = 0

v² = 6.25 m²/s²

v = 2.5 m/s

Now the quadratic equation will be as follows:

0.25 m/s² · t² - 2.5 m/s · t + 6.25 m = 0

Solving this equation:

t = 5 s

Remember that this is the time that takes the green car to catch the blue one after the green car arrived at the stop-light (5 s after it turned green). If we count from when the light turns green, the elapsed time will be 10 s.