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?
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
To find out the distance traveled by the object, we need to calculate the area under the velocity-time graph.
Since the graph is in the shape of a triangle, we can calculate the area by using the formula for the area of a triangle:
Area = (1/2) * base * height
In this case, the base of the triangle is the time it takes for the velocity to increase from 0 to its maximum value and then decrease back to 0, which is 10 seconds.
The height of the triangle is the maximum velocity reached by the object, which is 6 m/s.
Plugging these values into the formula, we get:
Area = (1/2) * 10 sec * 6 m/s = 30 m
Therefore, the object traveled a distance of 30 meters.