- #1
laneschmidt
- 4
- 1
This is a classic case of using the wrong formula for the question. My problem is that, using the given information, both formulas could've been used (to the best of my knowledge)
1. Homework Statement
A car traveling at 22.4 m/s comes to a stop in 2.55 s. Determine the stopping distance of the car, assuming that acceleration is uniform.
Let d=displacement
d=Vi*t+0.5at2
d=(Vi+Vf)/2t
a=∆v/t[/B]
1. I determined that the rate of acceleration was 8.78 m/s^2→ 22.4 m/s/2.55 s = 8.78 m/s2
2. I had all the variables needed so I used the above equation:
d=(22.4 m/s)(2.55 s) + 0.5(8.78 m/s2)(2.55)2
d=85.7 m (wrong answer)
3. After some confusion I decided to try another equation, the latter equation above:
d=(22.4 m/s + 0 m/s)/2(2.55 s)
d=28.6 m (right answer)[/B]
This confuses me. Clearly I do not know when to use each equation, even when both could be used with respect to the variables given. Some insight on this would be appreciated! Thank you.
1. Homework Statement
A car traveling at 22.4 m/s comes to a stop in 2.55 s. Determine the stopping distance of the car, assuming that acceleration is uniform.
Homework Equations
Let d=displacement
d=Vi*t+0.5at2
d=(Vi+Vf)/2t
a=∆v/t[/B]
The Attempt at a Solution
1. I determined that the rate of acceleration was 8.78 m/s^2→ 22.4 m/s/2.55 s = 8.78 m/s2
2. I had all the variables needed so I used the above equation:
d=(22.4 m/s)(2.55 s) + 0.5(8.78 m/s2)(2.55)2
d=85.7 m (wrong answer)
3. After some confusion I decided to try another equation, the latter equation above:
d=(22.4 m/s + 0 m/s)/2(2.55 s)
d=28.6 m (right answer)[/B]
This confuses me. Clearly I do not know when to use each equation, even when both could be used with respect to the variables given. Some insight on this would be appreciated! Thank you.