# A box sliding down a incline plane

1. Sep 20, 2009

### semc

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?

2. Sep 20, 2009

### RoyalCat

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
$$x(t)=x_0+v_0t+\tfrac{1}{2}at^2$$

Consider the case where the mass starts at rest, with $$x_0=0$$

3. Sep 20, 2009

### rl.bhat

Write down the relevant equations.

4. Sep 20, 2009

### semc

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

5. Sep 20, 2009

### Want to learn

x/(t*t) is the units of acceleration. For example meters per second squared

$$\frac {m/s^2}$$

6. Sep 20, 2009

### RoyalCat

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.

$$a(t)=constant$$
Taking the indefinite integral over time:

$$v(t)=a\cdot t + C$$

$$C=v_0$$

$$v(t)=a\cdot t+v_0$$

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

Taking the indefinite integral over time once more yields:

$$x(t)=\tfrac{1}{2}a\cdot t^2 +v_0\cdot t + C$$

$$C=x_0$$

$$x(t)=\tfrac{1}{2}a\cdot t^2 +v_0\cdot t + x_0$$

Q.E.D.

Just to elaborate on the point that has you confused though, differentiate the expression $$\tfrac{1}{2}at^2$$ with respect to time.
Remember, $$(ax^n)'=anx^{n-1}$$