1. ## Physics questions

Hello!

I have three physics questions.

1) if I have a velocity time graph, and the acceleration is not constant how can I calculate the displacement when the velocity is $35ms^-1$

2) A ball is dropped and let fall for 6 seconds until it crashes through a glass plate. In this crash it loses two thirds of its velocity. After the crash it takes the ball 2 s to reach the ground. How high from the ground is the glass plate?

3) How can I calculate the amount of days it takes for water with Radon with activity of 10 000 Bq/l to 1000 Bq/l, if the half-life of Radon is 3.8 days?

I feel hopeless. I don't know how to do these and the physics test is tomorrow. If someone could help me I would be really grateful.

Thank you!

2. Originally Posted by Coach
Hello!

Ok, I am just a beginner so sorry if I am wrong but hopefully I can type out the main bulk and then some more experianced people can confirm it. Sorry...

I have three physics questions.

1) if I have a velocity time graph, and the acceleration is not constant how can I calculate the displacement when the velocity is $35ms^-1$

I'm guessing it is a polygon. Just find the area of the space beneath the line, by splitting it up into different sections, this will equal the distance travelled (the displacement.)

If it is a curve, you will have to draw in your own polygon.

(I doubt it is a curve though because it would of just asked for an aproximation unless you are doing higher level where you would need to use more advanced maths.)

2) A ball is dropped and let fall for 6 seconds until it crashes through a glass plate. In this crash it loses two thirds of its velocity. After the crash it takes the ball 2 s to reach the ground. How high from the ground is the glass plate?

v = acceleration due to gravity (in this case) = 9.8
s = displacement = ?
t = time = 6
u = 0

s = (u+v)/2 x t

4.9 x 6 = 29.4

3) How can I calculate the amount of days it takes for water with Radon with activity of 10 000 Bq/l to 1000 Bq/l, if the half-life of Radon is 3.8 days?

10 000, 5000, 2500, 1250, 625

I think the answer is 3.8 x 4 = 15.2 days

This is for it to get to 625 though, maybe I have done this wrong though, haven't done it since gcse modulers.

I feel hopeless. I don't know how to do these and the physics test is tomorrow. If someone could help me I would be really grateful.

Thank you!
Hope this helps a little...

3. Originally Posted by Coach
1) if I have a velocity time graph, and the acceleration is not constant how can I calculate the displacement when the velocity is $35ms^-1$
The displacement will be the area between the velocity curve and the time axis. Note that areas underneath the time axis count as negative areas.

-Dan

4. Originally Posted by Coach
2) A ball is dropped and let fall for 6 seconds until it crashes through a glass plate. In this crash it loses two thirds of its velocity. After the crash it takes the ball 2 s to reach the ground. How high from the ground is the glass plate?
This goes in two parts.

Before the ball hits the glass:
I am using a coordinate system such that the origin is where the ball was dropped and the +y direction is upward. (This being the case, all positions for the ball will be negative.)

What we know:
$t = 6~s$
$y_0 = 0~m$ $y =$?
$v_0 = 0~m/s$ $v =$?
$a = -9.8~m/s^2$

We want both v and y, so to start:
$y = y_0 + v_0t + \frac{1}{2}at^2$

$y = \frac{1}{2}at^2$

$y = \frac{1}{2}(-9.8)(6)^2 = -176.4~m$

and
$v = v_0 + at = at = -9.8 \cdot 6 = -58.8~m/s$
(Again, the negative sign here merely indicates that the velocity is downward.)

Now for the second part. We know that the velocity of the ball just after it passes through the glass is $\frac{1}{3} \cdot -58.8 = -19.6~m/s$

I'm going to redefine the origin to be at the glass plate, and reset the clock so that it reads 0 s just after the ball passes through the glass. I am still taking +y to be upward.

Here's what we know:
$t = 2~s$
$y_0 = 0~m$ $y =$?
$v_0 = -19.6~m/s$ $v =$?
$a = -9.8~m/s^2$

To calculate y we again use:
$y = y_0 + v_0t + \frac{1}{2}at^2$

$y = \frac{1}{2}at^2$

$y = \frac{1}{2}(-9.8)(2)^2 = -19.6~m$

So the total height is (the negative of) the sum of these two displacements:
$h = 174.4 + 19.6 = 194~m$

-Dan

5. Originally Posted by Coach
3) How can I calculate the amount of days it takes for water with Radon with activity of 10 000 Bq/l to 1000 Bq/l, if the half-life of Radon is 3.8 days?
The "L" unit just indicates a density for the radon in the water. It plays no role in the Physics here.

A Bq indicates one disintegration per second, hence it is proportional to the amount of radioactive material present. So we may write the equation for the activity over time as:
$A = A_02^{-t/t^{1/2}}$
where A stands for the activity, $A_0$ stands for the activity at t = 0 days, and $t_{1/2}$ stands for the half-life. (I'm sorry, I can't recall the standard variable for activity.)

So we wish to solve
$1000 = 10000 \cdot 2^{-t/3.8}$
where t is measured in days.

$\frac{1000}{10000} = 2^{-t/3.8}$

$2^{-t/3.8} = \frac{1}{10}$

$-\frac{t}{3.8} = log_2 \left ( \frac{1}{10} \right )$

$-\frac{t}{3.8} = - log_2(10)$

$t = 3.8 \cdot log_2(10)$

I get $t = 12.6233~d$.

-Dan