Motion in a straight line tough one

  • #1
motion in a straight line....tough one

ok, a cyclist is travelling at 10ms^-1. he sees trafic lights ahead at 108m. He carries on cycling for a time T before decelerating and stopping at the lights 16.7seconds after he first saw the lights.

Calculate T

Im a bit stuck because i feel this question doesnt give enough information
If anyone could point me in the right way it would be much appreciated :)
 

Answers and Replies

  • #2
berkeman
Mentor
57,707
7,740


ok, a cyclist is travelling at 10ms^-1. he sees trafic lights ahead at 108m. He carries on cycling for a time T before decelerating and stopping at the lights 16.7seconds after he first saw the lights.

Calculate T

Im a bit stuck because i feel this question doesnt give enough information
If anyone could point me in the right way it would be much appreciated :)
I agree that it seems that you need more information to calculate T. The rate of deceleration would impact how long the constant velocity could be kept up.
 
  • #3


The question actually says
2 cyclists (B) travelling at 9ms^-1 and (A) travelling at 10ms^-1 are at the same point when they see traffic lights 108m away. B travels at its speed for a furthe 6seconds before decelerating and stopping at the lights. A travels for a time T before decelerating and reaching the lights at the same time as B

Calculate T

I worked out it took B 16.7 seconds to stop so figured that it has to be the same for A

Thats as much as i got :(
 
  • #4
berkeman
Mentor
57,707
7,740


The question actually says
2 cyclists (B) travelling at 9ms^-1 and (A) travelling at 10ms^-1 are at the same point when they see traffic lights 108m away. B travels at its speed for a furthe 6seconds before decelerating and stopping at the lights. A travels for a time T before decelerating and reaching the lights at the same time as B

Calculate T

I worked out it took B 16.7 seconds to stop so figured that it has to be the same for A

Thats as much as i got :(
Could you post your work for B that gave you 16.7 seconds to stop?
 
  • #5
1,434
2


Hm, maybe you have to assume they can only decelerate at a fixed number lambda, so that you could use the same number from B for A. I did get 12 seconds for B's stopping time though...

Anyway, do you have access to an answer to know you're right or not?
 
  • #6


6s*9ms^-1 = 54m
108m-54m = 54m

v^2=u^2+2as

0^2=9ms^-1 + 2a*54

a = -0.84359.....ms^-2

t=v-u/a

t=0ms^-1/0.8435...ms^-2
t = 10.7s

total time = 6 + 10.7 = 16.7s
 
  • #7


Hm, maybe you have to assume they can only decelerate at a fixed number lambda, so that you could use the same number from B for A. I did get 12 seconds for B's stopping time though...

Anyway, do you have access to an answer to know you're right or not?
Hi how did you get 12?

no access to answers :(
 
  • #8


Hi how did you get 12?

no access to answers :(
sorry yer got 12 now, me calculating wrong lol

couldnt i use T and work iit out algebraically in an equation?
 
  • #9
1,434
2


6s*9ms^-1 = 54m
108m-54m = 54m

v^2=u^2+2as

0^2=9ms^-1 + 2a*54

a = -0.84359.....ms^-2


t=v-u/a

t=0ms^-1/0.8435...ms^-2
t = 10.7s

total time = 6 + 10.7 = 16.7s
First of all you forgot the ^2 for the 9m/s, but that might be a typo. Still, if you have -81 = 108a, then a = 0.75

EDIT: woops, apparently I hadn't seen the last post before posting this...

And what do you mean "use T"? How can you use what you want to find? Anyway, if you can't assume A and B experience the same value for their acceleration, I think you simply don't have enough information to solve for T...
 

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