Asked by Olivia
The problem is: You apply your brakes to stop for a traffic light, and you come to a stop in a distance of 35 meters. How much time did it take you to stop?
from previous problems, i know:
INITIAL VELOCITY: 15 m/s
FINAL VELOCITY: 0 m/s
ACCELERATION: -3 m/s^2 (because you're slowing down)
DISPLACEMENT: 35 m
so we're solving for time.
i used the displacement= (initial velocity x delta t) + 1/2 (a)(delta t)^2 and got 3.7 seconds (used quadratic formula).
35= 15 (delta t) + 1/2 (-3) (delta t)^2
then put that into the quadratic formula.
[-15 +/- root 225-(4)(-1.5)(-35)]/-3
then (-15 +/- 3.87) / -3.
however, according to the answer sheet, the answer should be 4.7 seconds, not 3.7.. what did i do wrong?
from previous problems, i know:
INITIAL VELOCITY: 15 m/s
FINAL VELOCITY: 0 m/s
ACCELERATION: -3 m/s^2 (because you're slowing down)
DISPLACEMENT: 35 m
so we're solving for time.
i used the displacement= (initial velocity x delta t) + 1/2 (a)(delta t)^2 and got 3.7 seconds (used quadratic formula).
35= 15 (delta t) + 1/2 (-3) (delta t)^2
then put that into the quadratic formula.
[-15 +/- root 225-(4)(-1.5)(-35)]/-3
then (-15 +/- 3.87) / -3.
however, according to the answer sheet, the answer should be 4.7 seconds, not 3.7.. what did i do wrong?
Answers
Answered by
MathMate
INITIAL VELOCITY, v0: 15 m/s
FINAL VELOCITY, v1: 0 m/s
ACCELERATION, a: -3 m/s^2 (because you're slowing down)
DISPLACEMENT, S: 35 m
These numbers do not work.
The equation is supposed to be:
v1²-v0²=2aS
0-15² = 2*(-3)(35)
225 = 210
Something is amiss.
Correct either the acceleration or v0 will probably solve the problem.
FINAL VELOCITY, v1: 0 m/s
ACCELERATION, a: -3 m/s^2 (because you're slowing down)
DISPLACEMENT, S: 35 m
These numbers do not work.
The equation is supposed to be:
v1²-v0²=2aS
0-15² = 2*(-3)(35)
225 = 210
Something is amiss.
Correct either the acceleration or v0 will probably solve the problem.
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