so i got a question on a sheet for my physics homework, ill do the work myself ill just need some help with concept and formulas. so...
a motorbike ar rest starts to drive and after x amount of seconds it reaches a speed (about 150 kmph- remember i want to do this myself) at a certain distance (and no the distance and time/ displacement may be factors but are not what I'm looking for this is a question of acceleration) and the driver sees a rock in the distance at a certain distance away, assuming he has a reaction time of a certain amount of seconds and he slams the breaks and decelerates at twice the speed he was accelerating (ergo if he was accelerating at 10m/s-2 he would be decelerating at 20m/s-2, excuse any wrong terminology ) would he hit the rock in front of him?
please make this a formula and explain it so i have a better understanding.
a motorbike ar rest starts to drive and after x amount of seconds it reaches a speed (about 150 kmph- remember i want to do this myself) at a certain distance (and no the distance and time/ displacement may be factors but are not what I'm looking for this is a question of acceleration) and the driver sees a rock in the distance at a certain distance away, assuming he has a reaction time of a certain amount of seconds and he slams the breaks and decelerates at twice the speed he was accelerating (ergo if he was accelerating at 10m/s-2 he would be decelerating at 20m/s-2, excuse any wrong terminology ) would he hit the rock in front of him?
please make this a formula and explain it so i have a better understanding.
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It is a shame that you quote the speed as 150Km/hr.
If I explain how to convert this to m/s it will detract from the more important part of the question.
So I will replace it with a speed of 50m/s to keep things simple.
There are three parts to the movement.
At first he is accelerating.
Without needing to know the actual acceleration he will cover a distance given by average speed * time.
The average speed is half the maximum speed.
So once you know v = 50 and are given time = x then you can calculate the distance moved during this phase..
The distance moved in this part is s = v * x/2
In the next part of the movement he is travelling forwards at 50 m/s.
if the total time ( including reaction time ) is T then he moves 50 * T m during this phase.
Finally he brakes . As the acceleration is twice the original acceleration the time taken will be 1/2 x.
No matter what the initial acceleration.
so the distance moved is v/2 * 1/2 x = v * x/4
Now there is no one formula. There are three regions.
so total distance moved is
50*x/2 + 50 * T + 50 * x/4
= 50 * (3x/4 + T) where T is the time spent in the middle region. ( which must include the reaction time)
As long as this distance is less than the distance to the rock the motorcyclist will come to a safe stop.
If I explain how to convert this to m/s it will detract from the more important part of the question.
So I will replace it with a speed of 50m/s to keep things simple.
There are three parts to the movement.
At first he is accelerating.
Without needing to know the actual acceleration he will cover a distance given by average speed * time.
The average speed is half the maximum speed.
So once you know v = 50 and are given time = x then you can calculate the distance moved during this phase..
The distance moved in this part is s = v * x/2
In the next part of the movement he is travelling forwards at 50 m/s.
if the total time ( including reaction time ) is T then he moves 50 * T m during this phase.
Finally he brakes . As the acceleration is twice the original acceleration the time taken will be 1/2 x.
No matter what the initial acceleration.
so the distance moved is v/2 * 1/2 x = v * x/4
Now there is no one formula. There are three regions.
so total distance moved is
50*x/2 + 50 * T + 50 * x/4
= 50 * (3x/4 + T) where T is the time spent in the middle region. ( which must include the reaction time)
As long as this distance is less than the distance to the rock the motorcyclist will come to a safe stop.