The area-
The area under the line in a velocity-time graph represents the distance travelled. To find the distance travelled in the graph above, we need to find the area of the light-blue triangle and the dark-blue rectangle.
<span><span>Area of light-blue triangle -
<span>The width of the triangle is 4 seconds and the height is 8 meters per second. To find the area, you use the equation: <span>area of triangle = 1⁄2 × base × height </span><span>so the area of the light-blue triangle is 1⁄2 × 8 × 4 = 16m. </span></span></span><span> Area of dark-blue rectangle
The width of the rectangle is 6 seconds and the height is 8 meters per second. So the area is 8 × 6 = 48m.</span><span> Area under the whole graph
<span>The area of the light-blue triangle plus the area of the dark-blue rectangle is:16 + 48 = 64m.<span>This is the total area under the distance-time graph. This area represents the distance covered.</span></span></span></span>
Answer:
a. 11 m/s at 76° with respect to the original direction of the lighter car.
Explanation:
In this exercise, since both cars make a right angle, let's assume that the lighter car only has a horizontal velocity component (vx) and that the heavier one only has a vertical velocity component (vy). The final velocities for both components for the system can be determined as:
Assume that the lighter car has a 1kg mass and that the heavier car has a 4 kg mass.
The magnitude of the final velocity of the wreck can be found as:
![v_{f}^{2}= v_{fx}^{2}+ v_{fy}^{2}\\v_{f}=\sqrt[]{2.6^{2} + 10.4^{2}} \\v_{f}= 10.72](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%3D%20v_%7Bfx%7D%5E%7B2%7D%2B%20v_%7Bfy%7D%5E%7B2%7D%5C%5Cv_%7Bf%7D%3D%5Csqrt%5B%5D%7B2.6%5E%7B2%7D%20%2B%2010.4%5E%7B2%7D%7D%20%5C%5Cv_%7Bf%7D%3D%2010.72)
The final velocity has an intensity of roughly 11 m/s
As for the angle, it can be determined in respect to the lighter car (x axis) as follows:
Therefore, the wreck has a velocity with an intensity of 11 m/s at 76° with respect to the original direction of the lighter car.
I believe the correct
form of the energy function is:
u (x) = (3.00 N)
x + (1.00 N / m^2) x^3
or in simpler
terms without the units:
u (x) = 3 x +
x^3
Since the
highest degree is power of 3, therefore there are two roots or solutions of the
equation.
Since we are to
find for the positions x in which the force equal to zero, u (x) = 0,
therefore:
3 x + x^3 = u
(x)
3 x + x^3 = 0
Taking out x:
x (3 + x^2) = 0
So one of the
factors is x = 0.
Finding for the
other two factors, we divide the two sides by x and giving us:
x^2 + 3 = 0
x^2 = - 3
x = sqrt (- 3)
x = - 1.732 i, 1.732
i
The other two
roots are imaginary therefore the force is only equal to zero when the position
is also zero.
Answer:
x = 0
Answer:
frictonal force due to the surface of irregularities
