- #1

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I have always though to find the acceleration you can just use x/(t*t) but i use this formula but i couldn't get the same as the book. is there something wrong with my understanding?

- Thread starter semc
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- #1

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I have always though to find the acceleration you can just use x/(t*t) but i use this formula but i couldn't get the same as the book. is there something wrong with my understanding?

- #2

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You need to use the basic kinematic equation for motion under constant acceleration here. If you happen to need anything else, just differentiate it with respect to time

[tex]x(t)=x_0+v_0t+\tfrac{1}{2}at^2[/tex]

Consider the case where the mass starts at rest, with [tex]x_0=0[/tex]

- #3

rl.bhat

Homework Helper

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Write down the relevant equations.A 3kg block starts from rest at the top of a 30degree incline and slides a distance of 2m down the incline in 1.5s find the magnitude of the acceleration.

I have always though to find the acceleration you can just use x/(t*t) but i use this formula buti couldn't get the same as the book. is there something wrong with my understanding?

Show your calculations.

- #4

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can you explain why cant i just use x/(t*t) ?

You need to use the basic kinematic equation for motion under constant acceleration here. If you happen to need anything else, just differentiate it with respect to time

[tex]x(t)=x_0+v_0t+\tfrac{1}{2}at^2[/tex]

Consider the case where the mass starts at rest, with [tex]x_0=0[/tex]

- #5

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[tex] \frac {m/s^2} [/tex]

- #6

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It is not, however, the formula for the distance traveled. It may share the same units, but it's off by a factor of two.

[tex] \frac {m/s^2} [/tex]

Let's begin with the basics. We want to develop the kinematic equation for one dimensional motion under constant acceleration:

Our only givens are the initial position with respect to some origin, the initial velocities and accelerations with respect to a certain frame of reference, and the fact that the acceleration is constant.

[tex]a(t)=constant[/tex]

Taking the indefinite integral over time:

[tex]v(t)=a\cdot t + C[/tex]

[tex]C=v_0[/tex]

[tex]v(t)=a\cdot t+v_0[/tex]

We have now found an expression for the velocity as a function of time.

Taking the indefinite integral over time once more yields:

[tex]x(t)=\tfrac{1}{2}a\cdot t^2 +v_0\cdot t + C[/tex]

[tex]C=x_0[/tex]

[tex]x(t)=\tfrac{1}{2}a\cdot t^2 +v_0\cdot t + x_0[/tex]

Q.E.D.

Just to elaborate on the point that has you confused though, differentiate the expression [tex]\tfrac{1}{2}at^2[/tex] with respect to time.

Remember, [tex](ax^n)'=anx^{n-1}[/tex]

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