Asst_7_Week3Quiz2

#$&*

course PHY 201

2/25 8

Asst 7 Week 3 Quiz #2Determine the acceleration of an object whose velocity is initially 15 cm/s and which accelerates uniformly through a distance of 51 cm in 4.1 seconds.

Solution: v0= 15cm/s ds=51cm dt=4.1s Using ds= [(v0 +vf)/2] * dt, I solved for vf: 51cm = [(15cm/s + vf)/2] * 4.1s 51cm/4.1s = (15cm/s + vf)/2 2* (51cm / 4.1s) = 15cm/s + vf 24.88cm/s - 15cm/s = vf 9.88 cm/s = vf To solve for acceleration, I used: vf = v0 +a *dt and solved for a: 9.88cm/s = 15cm/s + a*(4.1s) -5.12cm/s = a(4.1s) -5.12cm/s / 4.1s = a -1.25cm/s^2 = a The object accelerates at about -1.25cm/s.

`gr31

Asst_7_Week3Quiz2

#$&*

course PHY 201

2/25 8

Asst 7 Week 3 Quiz #2Determine the acceleration of an object whose velocity is initially 15 cm/s and which accelerates uniformly through a distance of 51 cm in 4.1 seconds.

Solution:

v0= 15cm/s

ds=51cm

dt=4.1s

Using ds= [(v0 +vf)/2] * dt, I solved for vf:

51cm = [(15cm/s + vf)/2] * 4.1s

51cm/4.1s = (15cm/s + vf)/2

2* (51cm / 4.1s) = 15cm/s + vf

24.88cm/s - 15cm/s = vf

9.88 cm/s = vf

To solve for acceleration, I used: vf = v0 +a *dt and solved for a:

9.88cm/s = 15cm/s + a*(4.1s)

-5.12cm/s = a(4.1s)

-5.12cm/s / 4.1s = a

-1.25cm/s^2 = a

The object accelerates at about -1.25cm/s.

&#Good responses. Let me know if you have questions. &#