Question

The velocity of a particle moving along the -axis is given by f(t)=9-3t cm/sec. Use a graph of f(t) to find the exact change in position of the particle from time t=0 to t=4 seconds.
change in position =

is there anyone can help me please.. :(

Answer 1

f(t) = v(t) = ds/dt = change in position per unit time = 9-3t.

The position is the integral (studied them yet?) s = 9t -(3/2)t^2

The integral is the net area under your graph if you plot
f(t) = 9-3t vs t [y axis f(t), x-axis t]

Run a line from x=0, y=9 [0,9] to [3,0] the area in that triangle is the distance travelled in the first 3 seconds. Find the area with trig
A =(1/2) base times height = (1/2)(3)(9)
s = 27/2 cm

Continue that line into negative y territory to x=4, y=9-12=-3.

That little triangle is negative, and represents some backward motion, as f(t)<0. Calculate the area of that little triangle and subtract it from your 27/2 cm to get net distance in cm.

Answer 2

what would be the answer