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
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 answers
Thank you-I appreciate it