subtract from the 35 m the distance during reactiontime (v*.9) so the stopping distance is 35-.9v.
That is equal to d in ...
vf^2=v^2+2ad
vf is zero, solve for v.
What is the maximum speed at which a car could be moving and not hit a barrier 35.0 m ahead if the average acceleration during braking is -10.0 m/s2 and it takes the driver 0.90 s before he applies the brakes?
3 answers
35 - (.9)(-10.0) = 44
V = 29.664 and this is incorrect
Do I use the acceleration for the velocity or do I figure out the velocity from the informatin given?
V = 29.664 and this is incorrect
Do I use the acceleration for the velocity or do I figure out the velocity from the informatin given?
35 - (.9)(-10.0) = 44
V = 29.664 and this is incorrect
Do I use the acceleration for the velocity or do I figure out the velocity from the informatin given?
V = 29.664 and this is incorrect
Do I use the acceleration for the velocity or do I figure out the velocity from the informatin given?