Homework Help: Physics, Kinematics

Posted by Jackie on Thursday, December 1, 2011 at 8:06pm.

You drive with a constant speed of 15.1 m/s for 23.4 s. You then accelerate for 28.7 s to a speed of 25.8 m/s. You then slow to a stop in 37.1 s. How far have you traveled?

I've tried dividing this into 3 parts.
1st- constant speed
15.1m/s*23.4s=353.34 m

2nd-accelerating
I solved for acceleration, v=v0+at, a=0.0836 m/s^2, then plugged into x=x0+v0t+(1/2)at^2 where x=0+(23.4m/s)(28.7s)+(1/2)(.0836m/s^2)(28.7)^2=706.35

353.34+706.35=1059.35

I'm stuck after this. What do I do with the negative acceleration for the last part? I appreciate any guidance.

Physics, Kinematics - bobpursley, Thursday, December 1, 2011 at 8:28pm
ok on first leg.
Second leg, accelerating;
distance= average velocity*time
= 1/2 (15.1+28.7)*25.8

third leg:
distance= avg velocity*time
= 1/2 (28.7+0)*37.1
Physics, Kinematics - Jackie, Thursday, December 1, 2011 at 8:33pm
That isn't correct. Any other thoughts on this problem?
Physics, Kinematics - Jackie, Thursday, December 1, 2011 at 8:37pm
Nevermind, you just accidentally switched the velocity and time. It's (1/2)(15.1+25.8)*(28.7) and (1/2)(25.8)*37.1. Thank you so much for your help! I can't believe I didn't see that before.

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