The graph shows position versus time for an object moving with constant acceleration. What is the final velocity of the object at t = 10 s? Assume the initial velocity at the origin is 1 m/s.

The graph shows position versus time for an object moving with constant acceleration What is the final velocity of the object at t 10 s Assume the initial veloc class=

Respuesta :

Remark

The object is moving at a constant acceleration. The acceleration is not given, but it does permit you to use d = (vi + vt)*t/2

Givens

vi = 1 (This is important.

d = 21 m (read from the graph) You have to go up just a little bit from 20

vf = ???

t = 10 s

Solve

d = (vf + vi)*t/2

21 = (vf + 1) * 10/2 Multiply by 2

42 = (vf + 1) * 10 Divide by 10

4.2 = vf + 1 Subtract 1

4.2 - 1 = vf

vf = 3.2 Answer


The graph seems to show the position at t=10 s to be x=21 m. Thus, the position has changed at the average rate of

... (21-1) m/(10 s) = 2.0 m/s

The average is half the sum of initial and final velocities, so we have

... 2.0 m/s = (1/2)(1.0 m/s + vf)

... 4.0 m/s - 1.0 m/s = vf = 3.0 m/s

The final velocity is 3.0 m/s.