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A car is traveling along a straight road at a velocity of +36.0 m/s when its engine cuts out. For the next ten seconds, the car...Question
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 fifteen-second 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 ten-second interval.
Answers
drwls
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 = ?
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 = ?
Sara
31+ (-.5)(10)
26
Is that right?
26
Is that right?
drwls
Yes