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.

Question

How far did the object travel?

Answer 1

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

Answer 2

I don't get it. Explain more please

Answer 3

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.

Answer 4

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

Answer 5

Thank you. I get it now