Question

An object begins at an initial position of s = -2.0 m. The plot shows the object's velocity in the s-direction as a function of time. Find the object's: 

a. Displacement (Δs) at t = 9.0 s 

b. Position (s) at t = 9.0 s 

c. Average velocity (v_s,ave) between t = 0 s and t = 9.0s 

d. Acceleration (a) at t = 2.0, 6.0, and 8.0 s

Answer 1

Answer 2

Displacement in the given time is determined by calculating the area under the velocity time graph

a

b)

As initial position is -2,
Final position will be

c )

Average velocity is

d )

At t = 2,

At t= 6

At t= 8 s

Answer 3

Answer 4

SOLUTION :

a.

Displacement, ∆s

= Total area of given graph between 0 to 9 sec. above x- axis.

= 3 * 1 + (7 + 3) / 2 * 4 + 7 * 2 + (7 + 5) / 2 * 2

= 3 + 20 + 14 + 12

= 49 m (ANSWER).

b.

Position at 9.0 sec 

= s(0) + ∆s

= - 2.0 + 49

= 47 m (ANSWER).

c.

Average velocity in the interval 0 to 9 sec.

= Displacement / duration of interval

= 49 / 9 

= 5.44 m/s (ANSWER)

d.

           t (sec)                acceleration (m/s^2)

            2.0                            1 / 1 = 1.0

            6.0                                        0.0

            8.0                        - 1 / 1 = - 1.0