my ansewr :
about queation 8 I don't now how to solve : but i try
distant = 200 * 300 = 60000
b) dirction noth of west .
6.
a) Read the question well. It asks you for the final speed after breaking, not after accelerating from rest.
Can you try it again?
b) Uh... $\displaystyle \dfrac12(30)(10) = 150\ m$
But that's still not the answer. You didn't include the distance covered while braking.
If you draw a quick sketch of the vt graph of the motion of the car, you get the area of a triangle and a trapezium to find. Try this part again too.
7.
a) You didn't square the speed... wrong answer.
b) Where did you get that formula? The actual formula is $\displaystyle v = v_o + at$
What you are using doesn't exist and might be close to $\displaystyle v^2 = v_o\ ^2 + 2as$ but that's not it. Try again.
c) Uh... if it took t seconds to reach the highest point (obtained in part b), then it will take t seconds to fall down from the highest point... You don't need to calculate again.
Spoiler:
Or you can use $\displaystyle s = v_ot + \dfrac12 at^2$ where s = 0, t becomes easy to find. Oh, I find that you need to use this equation after reading the next part.
d) Now that you know the time, find the final speed using: $\displaystyle v = v_o + at$
8. For this one, if you don't know the cosine rule, you'll need to use Pythagoras' Theorem together with some trigonometric ratios.
Make a sketch. (This is very helpful in all sorts of problems)
When done, you'll find something like this:
You'll need to find x and y first, then use those values to find z, represented by the red broken line.
Then, you'll need to find the angle the red broken line makes with the horizontal line to get the direction C is from A.
6 ) I try agine now :
a)
final speed after breaking
mean we use a = - 4
v=v0+at = 0+(-4)(10) =-160 m/s^2
b) Dx = v^2-vo^2/2a
Dx = 3200
7
a ) = -31.8
b) v = v0 +at
0=25+-9.81
t = 2.55
c = It will take the ball the same amount of time to fall as it did to rise, 2.55
d = v=-v0
=-25
help me plese .
6 a) Okay... Did you make the sketch I told you to make?
You tell me yourself, from what you know about the gradients of lines. What is the final speed after the deceleration (at the end of the line with negative gradient. I made it to scale, so that the gradients, which are acceleration and deceleration show the exact situation.)
You reproduce it on a piece of paper if you feel like adding values that you find, but this is an easy way where you can confirm the values that you got through calculations. And I'll tell you again, sketches are easy way to do well without making mistakes.
b) Then, find the area under this graph, which is the distance covered.
7.
a) Um... I forgot to tell you also that you took the wrong symbol for the final and initial speed.
$\displaystyle v$ is the final speed of an object.
$\displaystyle v_o$ or $\displaystyle u$ is the initial speed of an object.
The ball is thrown with a speed of 25 m/s. Hence, the initial velocity is +25 m/s upwards (If you take upwards as positive). Then at its maximum height, the ball is temporarily at rest, hence, its final velocity is 0 m/s.
And if you think further, you'll see that you got -31.8 m which means the ball is 31.8 metres under the ground!!! (since you took the upwards direction as positive, negative means below)
b) Right.
c) Right, but just in case, I'll show you how to use the formula I gave you:
$\displaystyle s = v_ot + \dfrac12 at^2$
s is displacement, and in this case, since the ball hit the ground where you threw it, the displcament becomes 0.
vo is your initial velocity, +25m/s
a is your acceleration, -9.81 m/s^2
t is your time.
$\displaystyle 0 = 25t - \dfrac12 (9.81)t^2$
$\displaystyle 0 = t(25 - 4.905t)$
Solve for t. We get: t = 0 (that is before the ball is thrown and which makes sense)
We get:
$\displaystyle 25 - 4.905t = 0$
This gives: $\displaystyle t = 5.09684$
Since it takes 5.10 s for the whole 'trip' up and down, it takes (total time - time to go up) = 5.10 - 2.55 = 2.55 seconds to fall back down.
d) And you now see that it is with that same speed that you threw the ball, but only the direction changed.
8. I still need to see this one
Technically, it is wrong because the maximum height is not -31.8 m. The correct answer is 31.9 m (the value is 31.8877551m but rounded to 3 sf, we get 31.9 m)
$\displaystyle v^2 = v_o\ ^2 + 2as$
$\displaystyle 0 = (25)^2 + 2(-9.8)s$
$\displaystyle s = \dfrac{-625}{2(-9.8)} = 31.9 m$
The distance is right. and the direction is right
No, you see on the graph that the area is positive and hence, you cannot get a negative answer. I would suggest you label the velocity axis.
As the acceleration is 3 m/s^2, for each 1 second, the velocity increases by 3 m/s to that after 10 seconds, what speed is reached?
~~~~~~~~~~
If you consider formula-wise,
$\displaystyle x_1 + x_2 = v_1t + \dfrac12 at^2 + v_2t + \dfrac12 at^2$
You do not have the same initial speed. x1 is for the part where the slope is positive and x2 where the slope is negative. The initial speed at the beginning of this part is not zero from the graph as you can see.
You used the formula that I gave you in my previous post, or you find the area under the graph I gave you. but since you are not using the graph like I'm telling you, I'll do it for you.
USING THE GRAPH:
Since we know that the gradient of the first part is 3, at 10 seconds, the top speed is 10*3 = 30 m/s
The gradient of the second line is -4. In 6 seconds, the lost is speed is then 6*4 = 24 m/s
So, the speed after breaking becomes 30 - 24 = 6m/s
The area under the graph becomes:
Area of triangle + Area of trapezium = $\displaystyle \dfrac12 bh + \dfrac12 (l_1 + l_2)(h)$
$\displaystyle = \dfrac12 (10)(30) + \dfrac12 (30 + 6)(6)$
$\displaystyle = 150 + 108$
$\displaystyle = 258 m $
USING THE FORMULAE:
$\displaystyle x_1 + x_2 = v_1t + \dfrac12 at^2 + v_2t + \dfrac12 at^2$
$\displaystyle x_1 + x_2 = 0(10) + \dfrac12 (3)(10)^2 + (30)(6) + \dfrac12 (-4)(6)^2$
$\displaystyle x_1 + x_2 = 150 + 180 - 72$
$\displaystyle x_1 + x_2 = 258 m$
Same answers, different methods.