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Question

An object, which is at the origin at time t=0, has initial velocity v⃗ 0=(−14.0i^−7.0j^)m/s and constant acceleration a⃗ =(6.0i^+3.0j^)m/s^2

Find the position r⃗ where the object comes to rest (momentarily).
Express your answer in terms of the unit vectors i^ and j^.

Answer 1

vi = -14 i - 7 j m/s
a = 6i + 3j m/s

vf = 0 i + 0 j m/s

Find the time it takes to reach 0 m/s
vf = vi + a*t
t = vf - vi / a

in the i-direction
t = (0 m/s + 14 m/s) / 6 m/s^2 = 2.333 s

in the j-direction
t = (0 m/s + 7 m/s )/ 3 m/s^2 = 2.333 s

Since they both agree, t = 2.333 s

Equation of motion
s = s0 + vi*t + 1/2*a*t^2

In the i-direction
si = 0 - 14.0 m/s *2.333 s + 1/2*6 m/s^2 * (2.333 s)^2
si = -16.333 m

In the j-direction
sj = 0 - 7 m/s * 2.333 s + 1/2*3 m/s^2 * (2.333 s)^2
sj = -8.167 m

s = si + sj
s = -16.333 i - 8.167 j m

 

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