# Homework Help: What's the start speed and acceleration of the ball?

1. Sep 24, 2010

### Ockonal

1. The problem statement, all variables and given/known data
Using slant board, the ball was forced to move from bottom to the top. The ball was at the distance of 0.3 meters twice: after 1 and 2 seconds after start moving. What's the start speed and acceleration of the ball? (a=const)

2. Relevant equations
$$\newline x = x_0 + V_0*t + \frac{a*t^2}{2}$$
x₀ = 0 m
V₀ = 0 m/s
l = 0.3 m
t₁ = 1 s
t₂ = 2 s

3. The attempt at a solution
$$0.3 = 0 + V_0*1 + \frac{a*1^2}{2}$$

$$\newline 0.3 = 0 + V_0*2 + \frac{a*2^2}{2}$$
Here I can express, V₀. But I'm not sure the second expression is right. What's wrong?

2. Sep 24, 2010

### rl.bhat

Re: Acceleration

Hi Ockonal, welcome to PF.
Now subtract first equation from the second equation. You will get Vo in terms of a.
Put it in the first equation to find a. Substitute this value in the second equation to get Vo.

3. Sep 24, 2010

### Ockonal

Re: Acceleration

Oh, thanks.

4. Sep 24, 2010

### Ockonal

Re: Acceleration

Okay, I did the job:

$$0.3 = V_0 + \frac{a}{2}$$
$$0.3 = 2*V_0 + 2*a$$
=>
$$V_0 = 0.3 - \frac{a}{2}$$
$$0.3 = 2*(0.3 - \frac{a}{2}) + 2*a$$
$$0.3 = 0.6 + a;$$
$$a = -0.3 (m/s^2)$$
$$V_0 = 0.3 - \frac{a}{2};$$
$$V_0 = 0.15 (m/s)$$

The questions are:
Why the acceleration is negative? I think something is wrong in the first formulas.
The start speed is 0.45 m/s (refer to the answers), the acceleration is right (0.3)

5. Sep 25, 2010

### rl.bhat

Re: Acceleration

The acceleration is negative because the ball is moving up the inclined plane.
Vo = 0.3 - a/2 = 0.3 - (- 0.3/2) = 0.3 + 0.15 = 0.45 m/s

Last edited: Sep 25, 2010