Constant Acceleration Problem

[FaraDazed]

Homework Statement

A motorbike is riding along a straight road at 25m/s when the rider suddenly notices that there is a broken down car in his way, 65 meters ahead. He immediately brakes giving the bike a constant deceleration of 5m/s^2.

A. How far in front of the car does the biker come to a stop?
B. If the biker had not reacted a soon as he did and applied the brakes 1 second later, what speed would the bike had hit the car?

Homework Equations

v=u+at
v^2=u^2+2as
s=1/2(u+v)t
s=ut+1/2at^2

The Attempt at a Solution

The question confuses me a little as the question gives a distance of 65m in the question but asks you to also find a distance. But anyways, my attempts are below...

Part A:

v=0 a=-5 u=25

v^2=u^2+2as
0=625+(2x-5xs)
0=625+(-10xs)
s=(625-0)/-10

s=2.5mPart B:

v=0 t=1 a=-5 s=65

s=1/2(u+v)t
65=1/2(25+v)1
v=65/12.5

v=5.2m/s

 

[fornax]

Well I can tell you that you got Part a correct as far as my calculations, and here's why:
At 65m the cyclist applies the brakes and starts to decelerate at 5m/s^2. The problem then asks you how far before the car he stops, so his final velocity is 0. You obviously got that and your calculations brought you to 62.5m, so in turn 2.5 m in front of the car. It took him 62.5m to go from 25m/s to 0m/s, and since the car was at 65m, he didn't quite reach it. so 2.5m in front of the car. I hope that helps.

For part two:
Initially, before the brakes the cyclist is moving at 25m/s. At 65m he starts to apply the brake, but for this part, he doesn't. He waits a second, so he is still traveling 25m/s for another second, past that 65m mark.

 

[FaraDazed]

Well I can tell you that you got Part a correct as far as my calculations, and here's why:
At 65m the cyclist applies the brakes and starts to decelerate at 5m/s^2. The problem then asks you how far before the car he stops, so his final velocity is 0. You obviously got that and your calculations brought you to 62.5m, so in turn 2.5 m in front of the car. It took him 62.5m to go from 25m/s to 0m/s, and since the car was at 65m, he didn't quite reach it. so 2.5m in front of the car. I hope that helps.

For part two:
Initially, before the brakes the cyclist is moving at 25m/s. At 65m he starts to apply the brake, but for this part, he doesn't. He waits a second, so he is still traveling 25m/s for another second, past that 65m mark.

Thanks for the reply :)

Right I think I get the second part now, so if he waits one second, the distance between the bike and the car would now be 40m?

So part B would be..

v=25 s=40 a=-5 u=?

v^2=u^2+2as
625=u^2+(2x-5x40)
625=u^2+400
u^2=625-400
u^2=225
u=√225

u=15m/s ?

 

[fornax]

Looks good, that's exactly what I got for my answer as well. I'm not sure how you go about doing your problems, but I would highly suggest drawing a simple diagram of the problem at hand. Later on it is a good practice, and it keeps things orderly and harder to mix up. Even a line with some numbers on it might help clarify some things that may have been unclear before. Good luck with your future problems :)

 

[FaraDazed]

Edit: sorry realized my mistake.