thanks for everyone in this forum ,i am really getting lots of help here thanks alot regards, pease check attachments:D

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- May 22nd 2007, 11:16 PMcarlasaderA level Maths part3
thanks for everyone in this forum ,i am really getting lots of help here thanks alot regards, pease check attachments:D

- May 23rd 2007, 04:38 AMtopsquarkQuote:

1. A particle moves for 4 seconds in a straight line with uniform acceleration and describes 52 m. It then travels with uniform speed covering 48 m in 3 s. It is brought to rest by a retardation twice that of the initial acceleration.

*assume*that the motion is "smooth:" the final velocity of the initial phase is the same as the constant velocity of the second phase.

So assuming an origin at the point where the particle was when the time started and a +x direction in the direction of the initial acceleration we have:

$\displaystyle v = v_0 + at$

$\displaystyle v^2 = v_0^2 + 2a(x - x_0)$

where $\displaystyle v = \frac{48 \, m}{3 \, s} = 16 \, m/s$

So

$\displaystyle 16 = v_0 + 4a$

$\displaystyle 16^2 = v_0^2 + 2a(52)$

Solving the top equation for v_0 I get:

$\displaystyle v_0 = 16 - 4a$

Inserting this into the second equation gives:

$\displaystyle 256 = (16 - 4a)^2 + 104a$

Solving the quadratic for a I get that:

$\displaystyle a = -5.8503 \, m/s^2$ or $\displaystyle a = 2.3503 \, m/s^2$

We discard the negative solution since the acceleration is defined to be positive in the first phase, so $\displaystyle a = 2.3503 \, m/s^2$.

Thus $\displaystyle v_0 = 16 - 4(2.3503) = 6.59878 \, m/s$

The final retardation (acceleration) is just twice the negative of the acceleration in the first part so $\displaystyle a = -5.0606 \, m/s^2$.

You can use $\displaystyle v^2 = v_0^2 + 2a(x - x_0)$ to find the distance travelled in the last phase of the motion. I leave that to you.

-Dan - May 23rd 2007, 05:57 AMcarlasaderthanks
thanks alot Dan ,iwas wondering f u can help me with the rest of the problems ...i am stuck :rolleyes: :eek:

- May 23rd 2007, 06:16 AMtopsquarkQuote:

2. A stone is thrown vertically upward with a velocity of u meters per second. It passes a ledge in $\displaystyle t_1$ seconds and repasses it $\displaystyle t_2$ seconds after the start. Find the height of the ledge.

The stone's acceleration is -g, so the trajectory as a function of time will be

$\displaystyle y = y_0 + ut + \frac{1}{2}at^2$

$\displaystyle y = ut - \frac{1}{2}gt^2$

We know that the stone passes the ledge at two times, thus

$\displaystyle h = ut_1 - \frac{1}{2}gt_1^2$

$\displaystyle h = ut_2 - \frac{1}{2}gt_2^2$

Either of these expressions gives the height of the ledge.

-Dan - May 23rd 2007, 06:19 AMtopsquark
2.jpg is unreadable and 3.jpg is incomplete.

-Dan - May 23rd 2007, 06:30 AMcarlasaderthanks alot
thank you Dan , i tried opening it and it is readable ,i was wondering why it didnt open with u ,i think u just need to point the curser on it and maximize it i did that and it was readable ...thanks again Dan !!;)

- May 23rd 2007, 07:25 AMcarlasaderThanks!!! :)
I forgot to menton that:

Whenever a numerical value of g is required take g=9.8ms-2 (9.8 meters per second square).:eek: - May 23rd 2007, 11:32 AMtopsquarkQuote:

3. a)Express $\displaystyle 60 km h^{-1}$ in $\displaystyle m s^{-1}$.

$\displaystyle \frac{60 \, km}{1 \, h} \cdot \frac{1000 \, m}{

1 \, km} \cdot \frac{1 \, h}{3600 \, s} = 16.6667 \, m/s$

Quote:

3. b) Find the force which would bring a car of mass 800 kg to rest from a speed of 60 km/h in a distance of 50 m.

We need a value for a. Well, we know the change in velocity and the distance that acceleration acted over. So assuming a constant acceleration over this distance we get:

$\displaystyle v^2 = v_0^2 + 2a(x - x_0)$

I will assume an origin where car started to experience the force and a +x direction in the direction of the initial velocity of the car. Thus

$\displaystyle 0^2 = (16.66667)^2 + 2a(50)$

I get $\displaystyle a = 2.7778 \, m/s^2$. Thus

$\displaystyle F = (800 \, kg)(2.7778 \, m/s^2) = 2222.222 \, N$

-Dan - May 23rd 2007, 11:33 AMtopsquark
I would like to see just what you are having a problem with before I answer any more. That way I can help you better.

-Dan - May 24th 2007, 12:33 PMcarlasaderthanks alot
thank you Dan !!! ...well if u want to can continue in solving the others ,i am reviewing more when u explain ...thanks for all that :)