Question

The graph below plots velocity vs. time for a particle moving along a straight path. v (m/s) 16 12_ 18 27 36' (*) (a) What is the average acceleration of the particle during the following time intervals? (Enter your answers in m/s. Indicate the direction with the signs of your answers.) (i) Os to 9.00 S m/s2 (ii) 9.00 s to 27.0 s m/s2 (ili) Os to 36.0 s m/s2 (b) What is its instantaneous acceleration at each of the following times? (Enter your answers in m/s. Indicate the direction with the signs of your answers.) (i) 8.00 s m/s2 (ii) 18.05 m/s2 (III) 35.0 s m/s²

Answer 1

Part A) from 0 to 9 s

az - uf- vi -16 m/s - (-16 m/s) 9.0 s-Os a1 = 0 m/s2 di = =

9 to  27 s

Uf - Vi 16 m/s - (-16 m/s) 27.0 s-9 s 02 = 1.78 m/s2 (2=

0 to 36 s

ag=uf di 16 m/s - (-16 m/s) 36.0 s-Os az = 0.89 m/s2 (3 =

Part B) Instantaneous acceleration at 8.0 s

as velocity is constant.

a1 = 0 m/s

at t = 18 s

equal to the average acceleration between 9 to 27 as acceleration is constant.

a2 = 1.78 m/s

at t = 35.0 s

again as velocity is constant.

a3 = 0 m/s

Answer 2

ANSWER :

a.

Average acceleration between interval [0, 9] s 

= 0 m/s^2  (since velocity is constant). (ANSWER) 

Average acceleration between interval [9, 27] s  

= increase in velocity / Interval duration

= 32/18

= 1.78 m/s^2 (ANSWER)

Average acceleration between interval [0, 36] s 

= increase in velocity / Interval duration

= 32/36

= 0.89 m/s^2 (ANSWER)

b.

Instantaneous acceleration at time 8.00  s 

= Slope at time 8.00 s

= 0 m/s^2 (ANSWER).

Instantaneous acceleration at time 18.00  s 

= Slope at time 18.00 s

= 32/18 = 1.78 m/s^2 (ANSWER).

Instantaneous acceleration at time 35.00  s 

= Slope at time 35.00 s

= 0 m/s^2 (ANSWER).