A box sliding down a incline plane

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  • #1
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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 but i couldn't get the same as the book. is there something wrong with my understanding?
 

Answers and Replies

  • #2
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Your formula is incorrect.

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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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 but i couldn't get the same as the book. is there something wrong with my understanding?
Write down the relevant equations.
Show your calculations.
 
  • #4
346
2
Your formula is incorrect.

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]
can you explain why cant i just use x/(t*t) ?
 
  • #5
x/(t*t) is the units of acceleration. For example meters per second squared

[tex] \frac {m/s^2} [/tex]
 
  • #6
671
2
x/(t*t) is the units of acceleration. For example meters per second squared

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

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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