What's the start speed and acceleration of the ball?

[Ockonal]

Homework Statement

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)

Homework 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

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?

 

[rl.bhat]

Hi Ockonal, welcome to PF.
Your second expression is correct.
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.

 

[Ockonal]

Hi Ockonal, welcome to PF.
Your second expression is correct.
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.

Oh, thanks.

 

[Ockonal]

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)

 

[rl.bhat]

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