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You are to drive to an interview in another town, at a distance of 300 km on an expressway. The interview is at 11:15 a.m. You plan to drive at 100 km/h, so you leave at 8:00 a.m. to allow some extra time. You drive at that speed for the first 120 km, but then construction work forces you to slow to 42.0 km/h for 43.0 km. What would be the least speed needed for the rest of the trip to arrive in time for the interview?

Respuesta :

Answer:

[tex]v = 133.5 km/h[/tex]

Explanation:

As we know that the distance moved by the car is given 300 km

first it move for 120 km with speed

v = 100 km/h

so the time taken by car to move this distance is

[tex]t = \frac{d}{v}[/tex]

[tex]t = \frac{120}{100}[/tex]

[tex]t = 1.2 h[/tex]

Now it moves next 43.0 km with speed 42 km/h

so the time taken by it is given as

[tex]t = \frac{43}{42}[/tex]

[tex]t = 1.02 h[/tex]

now total time of journey is

[tex]t = 11:15 AM - 8 AM = 3.25 h[/tex]

time remaining to cover the last part of journey is

[tex]\Delta t = 3.25 - (1.02 + 1.2) = 1.026 h[/tex]

now the remaining distance is

[tex]d = 300 - 120 - 43 = 137 km[/tex]

so the speed of the car is given as

[tex]v = \frac{137}{1.026}[/tex]

[tex]v = 133.5 km/h[/tex]

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