Posted by Sara on Sunday, November 30, 2008 at 7:27pm.
A car is traveling along a straight road at a velocity of +31.0 m/s when its engine cuts out. For the next ten seconds, the car slows down further, and its average acceleration is a1. For the next five seconds, the car slows down further, and its average acceleration is a2. The velocity of the car at the end of the fifteensecond period is +24.5 m/s. The ratio of the average acceleration values is a1/a2 = 1.67. Find the velocity of the car at the end of the initial tensecond interval.

Physics  drwls, Sunday, November 30, 2008 at 7:48pm
OK, so a1 = 1.67 a2
After 15 seconds, the velocity is
31 + a1*10 + a2*5
= 31 + 1.67 a2 *10 + a2* 5
= 31 + 21.7 a2 = 24.5
21.7 a2 = 6.5
a2 = 0.30 m/s^2
a1 = 0.50 m/s^2
V(@ 10 s) = V + a1*10 = ? 
Physics  Sara, Sunday, November 30, 2008 at 8:54pm
31+ (.5)(10)
26
Is that right? 
Physics  drwls, Sunday, November 30, 2008 at 9:49pm
Yes