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
Displacement, Velocity and Acceleration
In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.
Linear Displacement
The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F84f39a0e-1c16-43ec-bcb3-7f889ca233a2%2F69892538-331a-4145-8f68-8f4a2403094b%2Fp0ao1zh_processed.png&w=3840&q=75)

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