The velocity vs time graph of an object is approximated by a triangle which starts at v=0 at t=0, rises to a maximum of v=6 m/s at t=6 sec, then returns to zero at t=10 sec.

How far did the object travel?

5 answers

distance is the integral of the velocity. You have defined v(t) as a piecewise function.

v(t) =
t for 0<=t<=6
6-(3/2)(t-6) for 6<=t<=10

the distance is thus
∫[0,6] t dt + ∫[6,10] 15-(3/2)t dt
I don't get it. Explain more please
if you have not studied calculus yet, what methods have you studied that relate distance to speed? There must be similar problems in your text.
ok - I'll assume that that you have had some physics exercises where you have learned that with acceleration a and initial velocity v,

s = vt + 1/2 at^2

Since v went from 0 to 6 in 6 seconds, a = 6/6 = 1 m/s^2

So, to get the distance during acceleration, you have

s = 0*6 + 1/2 (6^2) = 18 m

During deceleration, a = -3/2 m/s^2 and the distance traveled is

s = 6*4 - (3/4)*4^2 = 12 m

so, total distance is 30m
Thank you. I get it now