When James applied the brakes to his car, it slows unifromly from 15.0 m/s to 0.0 m/s in 2.50 s. How many meters before a stop sign must she apply her brakes in order to stop at the sign?
Answers:
6.0m
37.5 m
18.8 m
48.0m
Please help-I've just started Physics last week and I'm already screwed up with which formular to use-Could you please explain?
Thank you
2) second grade formula:
distance= averageveloicty*time
= (1/2)(15+0)*2.5
3) hard way:
Vf^2=Vi^2+2ad
where a is changevelocity/time=(0-15)/2.; and notice a is negative.
and solve for d.
4) still harder: compute a as above.
distance=Vi*t+1/2 a t^2
Thank you-I appreciate it
To solve this problem, you need to use the equation of motion for uniformly decelerating objects:
v^2 = u^2 + 2as,
where v is the final velocity (0.0 m/s), u is the initial velocity (15.0 m/s), a is the acceleration (which we will assume to be constant), and s is the distance traveled.
Since we want to find the distance before the stop sign where James should apply brakes, we can rearrange the equation to solve for s:
s = (v^2 - u^2) / (2a).
Now, let's substitute the given values into the equation:
u = 15.0 m/s,
v = 0.0 m/s,
a = ?,
s = ?.
We are given the time taken (2.50 s), which we can use to find the acceleration by rearranging the equation:
a = (v - u) / t.
Substituting the values:
a = (0.0 m/s - 15.0 m/s) / 2.50 s.
Now, we can substitute the value of acceleration into the equation for distance:
s = (0.0 m/s)^2 - (15.0 m/s)^2 / (2 * [(0.0 m/s - 15.0 m/s) / 2.50 s]).
Simplifying:
s = -(15.0 m/s)^2 / (2 * (-15.0 m/s / 2.50 s)).
Now, evaluate the expression:
s = -(225.0 m^2/s^2) / (-30.0 m/s^2/s).
The units in the denominator cancel out, leaving us with meters:
s = 7.5 m.
Therefore, James must apply her brakes 7.5 meters before the stop sign in order to stop at the sign.
None of the provided options matches the correct answer.