Ask your own question, for FREE!
Physics 13 Online
OpenStudy (anonymous):

A car accelerates from rest at a constant rate of 2.0m/s^2 for 5.0s. (a) What is the speed of the car at the end of that time? (b) How far does the car travel in this time? I'm having quite a bit of trouble with these kinematic formulas and their use and would appreciate any help you might have. Thank you. Using s=v^2/2a I get 25m but I'm not sure if that is correct.

OpenStudy (anonymous):

your answer for the distance traveled is correct final position = initial position + initial velocity*time + 0.5*acceleration*time^2 final position = 0 + 0*5 + 0.5(2 m/s^2)*(25 s^2) = 25

OpenStudy (ghazi):

V=U+a*t V=2*5=10m/s distance traveled=10*5=

OpenStudy (ghazi):

also S=0.5*2*(5)^2=0.5*2*25

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!