# A-Level Kinematics tricky question

hi, can someone help with this question please? It looked like it would be easy at first but I can't get the right answer. Thanks

Q: A car travelling with constant acceleration is timed to take 15s over 200m and 10s over the next 200m. What is the speed of the car at the end of the observed motion?
Original post by DFranklin

Actually it's alright now, I've solved it after looking through earlier parts of the textbook. I was assuming u=0 even though it didn't specifically say so. Anyway, the answer was 22.66ms^-1

200=15u+112.5a
400=25u+312.5a

1000=75u+562.5a
1200=75u+938.5a
-200=-375a
a=0.5333....

200=15u+(112.5*0.533)
200=15u+60
u=28/3

v^2=(28/3)^2+2(0.5333...x400)
v^2=513.8
v=22.66ms^-1
Original post by hard-working-pen
hi, can someone help with this question please? It looked like it would be easy at first but I can't get the right answer. Thanks

Q: A car travelling with constant acceleration is timed to take 15s over 200m and 10s over the next 200m. What is the speed of the car at the end of the observed motion?

The answer is 22.7m/s to 3 sigfig