# Mechanics

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Thread starter 3 months ago
#1
"10. A car is behind a tractor on a single-lane straight road with one lane in each direction. Both are moving at 15 m s^-1. The speed limit is 25 m s^-1, so the car wants to overtake. The safe distance between the car and the tractor is 20 m

a) To overtake, the car goes onto the other side of the road and accelerates at a constant rate or 2 m s^-2 until reaching the speed limit, when it continues at a constant speed. Show that the distance the car is ahead of the tractor at time t s after it starts to accelerate is given by t^2 - 20 for 0 < t =< 5, and deduce that the car is not a safe distance ahead of the tractor before reaching the speed limit.

b) The car pulls in ahead of the tractor once it is a safe distance ahead. Find the total time taken from the start of the maneuver until the car has safely overtaken the tractor."

could somebody please help me by breaking this down, i've been sat at it forever now and i'm still not sure how to get to the correct answer.
0
3 months ago
#2
So for the first part of the question you know that the car starts 20 metres behind the tractor. You also know that they are starting with the same velocity. You then know that the car is going to accelerate until the speed limit, at 2m/s^2, which means it will accelerate for 5 seconds before continuing at this constant velocity. For part a), you are calculating how far the car is ahead of the tractor after this period of acceleration. You need 2 suvat equations here, one for distance and one for velocity- s = ut + 1/2(a) (t^2) and v = u + at

a) the distance the car is ahead after a time, t, is D = s(car) - s(tractor) - 20. The -20 comes from the fact the tractor starts 20m ahead of the car.
s(car) = 15t + 1/2(2)(t^2)
s(tractor) = 15t

therefore D = [15t + 1/2(2)(t^2)] - [15t] - 20 = t^2 - 20

In order for the car to be a safe distance ahead after 5 seconds, it needs to be 20 metres ahead of the tractor. To find out how far ahead the car is after 5 seconds, we plug in t = 5 to our equation for D----> D = 5^2 -20 = 25-20 = 5m.

So no it is not a safe distance ahead

b) We know for the car to be a safe distance ahead, it has to be 20m ahead. From part a) we know that after 5 seconds, the car is 5m ahead. This means the car still has another 15m to gain on the tractor to be at a safe distance. However, we know that the car has reached the speed limit and is no longer accelerating, which means that v(car) = 25m/s and v(tractor) = 15m/s.

To work out how long it takes for the car to gain 15 m, we use our equation s = ut (distance = speed * time).
The difference in distance needs to be s = s(car) - s(tractor) = 15m.
s(car) = 25t
s(tractor) = 15t

therefore:
15 = 25t-15t
15 = 10 t
t = 1.5 seconds

This means the total time for the manoeuvre is t(total) = 5s + 1.5 s = 6.5 s

I hope this was helpful, let me know if there's anything you're still confused about
Last edited by Randomuser12343; 3 months ago
1
Thread starter 3 months ago
#3
thank you so so much, this helped me in the right direction, i overlooked all of this completely
0
2 weeks ago
#4
Hi, I need help with another part of the question for the same situation.

c) To overtake safely on the single-lane road, when the car returns to the correct side of the road in front of the tractor there must be a gap between the car and oncoming traffic of at least 20 m. Assuming a car travelling in the opposite direction is moving at the speed limit, find the minimum distance it must be from the initial position of the overtaking car at the point at which it starts to overtake.

Hope there is anyone that can help me with this question. Thank you.
Last edited by ewchiew; 2 weeks ago
0
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