Physics velocity Question!?
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Physics velocity Question!?

[From: ] [author: ] [Date: 14-02-19] [Hit: ]
10t + 4 = -1.0 = -1.t = 0.961 sANS a.Vcommon = 10(0.961) + 4 = 13.......
Two objects are moving along the x axis. The position of object 1 is given by x = 5t2 + 4t + 2 (m/s). The acceleration of object 2 is given by a = -3t (m/s2) and its velocity is +15 m/s at 0s.
a. When do both objects have the same velocity?
b. Refer to (a), what is the velocity?

-
obj 1: x = 5t² + 4t + 2
obj 1: dx/dt = V1 = 10t + 4

obj 2: a = -3t
obj 2: V2 = integral -3t = -3t²/2 + C
at t = 0 Vo = C = + 15 m/s
obj 2: V2 = -1.5t² + 15
V1 = V2 at t = ?
10t + 4 = -1.5t² + 15
0 = -1.5t² -10t + 11
solve quadratic for positive t value:
t = 0.961 s ANS a.
Vcommon = 10(0.961) + 4 = 13.6 m/s ANS b.
chk on ANS b.
Vcommon = -1.5t² + 15 = 13.61 m/s √
1
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