# help on constant acceleration question

#1
A car is being driven on a straight stretch of motorway at a constant velocity of 34 m/s when it passes a velocity restriction sign S warning of road works ahead and requiring speeds to be reduced to 22 m/s. The driver continues at her velocity for 2s after passing S. She then reduces her velocity to 22 m/s with constant deceleration of 3 m/s^2, and continues at the lower velocity.

Find the shortest distance before the road works that S should be placed on the road to ensure that a car driven in this way has had its velocity reduced to 22 m/s by the time it reaches the start of the road works.
0
4 years ago
#2
(Original post by RickHendricks)
A car is being driven on a straight stretch of motorway at a constant velocity of 34 m/s when it passes a velocity restriction sign S warning of road works ahead and requiring speeds to be reduced to 22 m/s. The driver continues at her velocity for 2s after passing S. She then reduces her velocity to 22 m/s with constant deceleration of 3 m/s^2, and continues at the lower velocity.

Find the shortest distance before the road works that S should be placed on the road to ensure that a car driven in this way has had its velocity reduced to 22 m/s by the time it reaches the start of the road works.
Determine the amount of time it takes after passing S to get down to the velocity of 22m/s.

Determine the distance travelled in this time. This will be your answer.

If you're struggling, sketch a velocity-time graph for the motion during the time when the car passes S, and until the velocity is reduced to 22 m/s. You want to find the area under this graph, which is the distance travelled.
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#3
(Original post by RDKGames)
Determine the amount of time it takes after passing S to get down to the velocity of 22m/s.

Determine the distance travelled in this time. This will be your answer.
We haven't been given any information though. If there was a speed limit on the motorway, then we would know what the greatest velocity it would be theoretically. however the question does not specify what it means exactly by "this way".

Is it meaning if all the cars imitate the exact motion as the before one, or all the cars coming on the road, that direction.

It's basically asking us to work out the distance from the roadworks we would have to put a sign to ensue that ALL cars CAN decelerate to 22 m/s BEFORE reaching the road works. It also has not specified in the actual context whether it has reached the road works exactly when it drops down to 22 m/s or the car then continues on travelling at that 22 m/s before reaching the road works? If the second bit was true, then shouldn't we be given something like "it travels at this speed for another x s before reaching the road works".

The question is really confusing me..., the mark scheme says 180, but unfortunately it doesn't state the methods.
0
4 years ago
#4
(Original post by RickHendricks)
...
You're reading into this too much.

It's looking for the answer specific to this one car.
1
#5
(Original post by RDKGames)
You're reading into this too much.

It's looking for the answer specific to this one car.
so that does mean that we can assume that U = 34? and that V = 22? even so, if we use the fact that we know the acceleration is -3, we would still get a distance of 112, which is the wrong answer.

This is the V-T graph.
0
4 years ago
#6
(Original post by RickHendricks)
so that does mean that we can assume that U = 34? and that V = 22? even so, if we use the fact that we know the acceleration is -3, we would still get a distance of 112, which is the wrong answer.

This is the V-T graph.
No we don't. The graph stops at at some value which you need to determine first by using . Find the area under this graph, meaning all the way down to the axis.
0
1 year ago
#7
hey I know this is three years late lol but by any chance does anyone know why using suvat --> (22)^2 = 34^2 +2(-3)Y gives 112m which is the wrong answer?
Last edited by username3477548; 1 year ago
0
1 year ago
#8
(Original post by Qxi.xli)
hey I know this is three years late lol but by any chance does anyone know why using suvat --> (22)^2 = 34^2 +2(-3)Y gives 112m which is the wrong answer?
youve not added on the 68 for the 2s motion at the start
1
1 year ago
#9
(Original post by mqb2766)
youve not added on the 68 for the 2s motion at the start
ohh I see I didn't realize we had to add that on
thank you!! x
0
1 year ago
#10
(Original post by Qxi.xli)
ohh I see I didn't realize we had to add that on
thank you!! x
From gcse physics, it would be the "thinking time" and youre working out the total stopping distance as
thinking distance + braking distance.
Its always worth underlining / sketching each fact in the question and making sure you don't miss anything out.
0
1 year ago
#11
(Original post by mqb2766)
From gcse physics, it would be the "thinking time" and youre working out the total stopping distance as
thinking distance + braking distance.
Its always worth underlining / sketching each fact in the question and making sure you don't miss anything out.
ooh yeah that makes sense
thanks
0
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