Help find instantaneous velocity at X seconds?

v = ds/dt so calculate the slope at each point:

t=0.5 s:

v = (10 - (-3)) m / 1s = 13 m/s

t=2.5 s:

v = (5 - 10)m / (4-1)s = -5/3 m/s = -1.67 m/s

t=4.5 s:

v = 0 (horizontal line= 0 slope, well spotted :)

t=8.5 s:

v = (0 - (-5))m / 1s = 5 m/s

[the first answerer is incorrect at t=2.5s]

The instantaneous velocity at any time is the slope of the position-velocity point at that time. Since in all cases, the graph is a straight line segment, the slope is found by finding the position at any two time points on that segment, subtract them and divide by the time difference. (Subtract the early point from the later one.)

At 0.5 s: Position at 0 s = -3 m, position at 1 s = 10 m. The difference is 10 - (-3) = 13 m; the time difference is 1 - 0 = 1 s, so the velocity is 13 m/s

At 2.5 s: Position at 2 s = -5 m, position at 9 s = 0 m. The difference is 0 - 8 = -3 m; the time difference is 4 - 2 = 2 s, so the velocity is -1.5 m/s

At 4.5 s: Position at 4 s = 5 m, position at 5 s = 5 m. The difference is 5 - 5 = 0 m; , so the velocity is 0 m/s

At 8.5 s: Position at 8 s = -5 m, position at 9 s = 0 m. The difference is 0 - (-5) = 5 m; the time difference is 1 s so the velocity is 5 m/s