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An object traveling along the x axis at constant acceleration has a velocity of +10 m/s when it is at x= 6.0 m and of +15 m/s when it is at x = 10 m, what is its acceleration? Ans: 16m/s²

User Txs
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Final answer:

The acceleration of an object traveling along the x-axis from x=6.0 m to x=10 m, with a change in velocity from +10 m/s to +15 m/s, is calculated using a kinematic equation. It is found to be 15.625 m/s², which conflicts with the student's answer of 16 m/s².

Step-by-step explanation:

The question is asking us to determine the acceleration of an object that is moving along the x-axis with varying velocities at different positions. To find the acceleration, we can use the kinematic equation:

v^2 = u^2 + 2as,

where

v is the final velocity,

u is the initial velocity,

a is the acceleration, and

s is the displacement.

Given that the velocity changes from +10 m/s to +15 m/s and the displacement (x) changes from 6.0 m to 10 m, the values are:

u = +10 m/s (initial velocity),

v = +15 m/s (final velocity),

s = 10 m - 6 m = 4 m (displacement).

Plugging these into the equation, we get:

(+15 m/s)^2 = (+10 m/s)^2 + 2 * a * 4 m,

225 m^2/s^2 = 100 m^2/s^2 + 8a m/s^2,

125 m^2/s^2 = 8a m/s^2,

a = 125 m^2/s^2 / 8 m/s^2,

a = 15.625 m/s^2.

The calculated acceleration is 15.625 m/s^2, rather than the answer provided by the student which is 16 m/s^2. It could be a typing error or a rounding off.

User An Employee
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