Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

An electric vehicle starts from rest and accelerates at a rate of 2.0 m/s2 in a straight line until it reaches a speed of 20 m/s. The vehicle then slows at a constant rate of 1.0 m/s2 until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle travel from start to stop?

OpenStudy (anonymous):

(a) total time elapsed =10+20 =30s

OpenStudy (anonymous):

total distance travelled =100 +200 =300 m

sam (.sam.):

v = u + at 20 = 0 + 2t1 t1 = 10 sec 2nd phase:- v = u - at 0 = 20 - 1 x t2 t2 = 20 sec total time = t1+t2 = 10 + 20 = 30 sec (b) 1st phase: s = ut + 1/2at^2 s1 = 0 + 1/2 x 2 x (10)^2 s1 = 100 m 2nd phase: By s = ut - 1/2at^2 s2 = 20 x (20) - 1/2 x 1 x (20)^2 s2 = 200 m total distance(s) = s1+s2 = 100 + 200 = 300 m

OpenStudy (anonymous):

woohoo. you are legend!

OpenStudy (anonymous):

sorry matricked but no working...no medal...i got the first one already though... the second one confused me

OpenStudy (anonymous):

don't hate the player...hate the game...

OpenStudy (anonymous):

see for calculating time use velocity/acc for both retardation and acc as the vehicle starts from and stops to 0. for distance use distance= average velocit * time and in each case its (20+0)/2 thus its easy for others to do orally

OpenStudy (anonymous):

average velocity * time is used whenever acc. or retardation is constant ... it helps in getting the answers quickly it time is known

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!