Question

Please help!
A physics student gets on a bike. Not knowing that the handlebars are locked, he goes speeding away and comes to an intersection where there is a red light. With the maximum deceleration of 2.00 m/s^2 and a distance of 54.5 m to the intersection, what is the maximum velocity that he can be going and still stop at the intersection?

I'm not sure which equation to use for this problem. I've worked this problem out many times and come up with different answers.
I've come up with 10.44 m/sec and 14.76 m/sec. Please help me understand this problem.

Answer 1

i think its 14.76 m/s. the logic is final speed is zero and initial ie. @ 54.5m from intersection we have to calculate. so just use Newton's equation v^2 = u^2 + 2*a*s. v: final speed, u: initial speed. s: displacement and a: acceleratio/ decceleration.

Answer 2

But with significant figures, would it be 14.76 or just 14.8?

Answer 3

First look at what you know:
1. The total distance that will be traveled (s) = 54.5 m
2. The acceleration (a) = -2.0 m/s^2 (negative due to accel opposite motion)
3. The final velocity (vf) = 0 (we know he comes to a stop)
So we need the initial velocity (vi) = the velocity max he can have and stop in the distance based on the given deceleration.

(vf)^2 = (vi)^2 + 2as

*** Note that you must rearrange equation and remember (a) is negative, to solve for (vi).

Answer 4

it comes as square root (218). well i think you can write it as 14.8m/s.

Answer 5

0^2 = (vi)^2 + 2(-2.00) * 54.5
(vi)^2/-4.00 = 54.5
(vi)^2 = -218
(vi) = 14.76 = 14.8

Answer 6

Seem about right?

Answer 7

Thank you both for your help! I appreciate it!

Answer 8

I'm going to close this one soon because I have another question I need answered!

Answer 9

You are welcome. Since I teach and tutor I just try to show method of thinking not just give answer.

Answer 10

I appreciate your tutoring aspect.

Answer 11

even i do. @SWdrafter