An object moving with uniform acceleration has a velocity of 11.5 cm/s in the positive x-direction when its x-coordinate is 3.05 cm. If its x-coordinate 3.00 s later is -5.00 cm, what is its acceleration? Step 1 The object has moved to a particular coordinate in the positive x-direction with a certain velocity and constant acceleration; then it reverses its direction and moves in the negative x-direction to a particular x-coordinate in time t. We are given an initial velocity v₁ = 11.5 cm/s in the positive x-direction when the initial position is x, = 3.05 cm (t = 0). We are given that at t = 3.00 s, the final position is x = -5.00 cm. The acceleration is uniform so that we have the following equation in terms of the constant acceleration a. xf = x₁ = V₁t + 11at² Now we substitute the given values into this equation. cm)-([ | cm) = m/s) ([ ₁) + 1/²([ S S

Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
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Chapter2: One Dimensional Motion
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An object moving with uniform acceleration has a velocity of 11.5 cm/s in the positive x-direction when its x-coordinate is 3.05 cm. If
its x-coordinate 3.00 s later is -5.00 cm, what is its acceleration?
Step 1
The object has moved to a particular coordinate in the positive x-direction with a certain velocity and constant acceleration; then it
reverses its direction and moves in the negative x-direction to a particular x-coordinate in time t. We are given an initial velocity
v; = 11.5 cm/s in the positive x-direction when the initial position is x; = 3.05 cm (t = 0). We are given that at t = 3.00 s, the final
position is x = -5.00 cm. The acceleration is uniform so that we have the following equation in terms of the constant acceleration a.
xf = x₁ = V₁t + at²
Now we substitute the given values into this equation.
cm)-(C
cm) = (
15) ([
cm/s
] ₁) + 12/2² (
s)²
Transcribed Image Text:An object moving with uniform acceleration has a velocity of 11.5 cm/s in the positive x-direction when its x-coordinate is 3.05 cm. If its x-coordinate 3.00 s later is -5.00 cm, what is its acceleration? Step 1 The object has moved to a particular coordinate in the positive x-direction with a certain velocity and constant acceleration; then it reverses its direction and moves in the negative x-direction to a particular x-coordinate in time t. We are given an initial velocity v; = 11.5 cm/s in the positive x-direction when the initial position is x; = 3.05 cm (t = 0). We are given that at t = 3.00 s, the final position is x = -5.00 cm. The acceleration is uniform so that we have the following equation in terms of the constant acceleration a. xf = x₁ = V₁t + at² Now we substitute the given values into this equation. cm)-(C cm) = ( 15) ([ cm/s ] ₁) + 12/2² ( s)²
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