Acceleration
Quiz
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Some parks have rides slowly carrying up people to the top before abruptly dropping down, increasing everyone's velocity.
An aircraft would is greater thrill, weighting 35,000 kg jets from to 250 km/h in 2.5 seconds.
Objects changing velocity is acceleration; how fast an object's velocity changes overtime, which can be shown by the velocity-time graph, like the position-time graph; time is always the x-axis and velocity is the y-axis, instead of position.
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acceleration (aav):
Velocity-time graphs of both uniform velocity and uniform acceleration are always straight lines. The position-time graph of an accelerated motion is curved.
uniform: remaining the same
Determining acceleration from a velocity-time graph
Determining acceleration from a velocity-time graph
Figure 1 shows a velocity-time graph for skateboard rolling down a ramp. See that the graph's line goes ascends to the right and has an x and y-intercept of zero.
The graph's slope of Figure 2 is calculated using the standard slope equation.
slope = rise/run
Figure 1
Recall that SI unit for velocity is m/s and the SI for time is seconds. The unit for acceleration is a derived unit, created by combining SI base units. We can derive units for acceleration by diving a velcoity unit (m/s) by a time units:
m/s m 1 m
___ = ___ x __ = __
s x s s2
"square seconds": m/s2 isn't saying that it's measured a square second. It's a shorter way to show the derived way. It's also be read as "metres per second", showing how many metres per velocity second are gained or lost per acceleration second.
Sub problem 1
Sub problem 1
What's the acceleration be if the same velocity change of 30 m/s took place over a time interval of only 5 s?
Solution:
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aav = 30 / 5
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aav = 6 m/s2
See how as the time interval decrease, it increases the acceleration rate.
Problem 1
Problem 1
What's the skateboard's acceleration in the motion between 0 s and 10 s in Figure 1.
Solution:
At 0 seconds is 0 velocity and at 10 seconds is 30 velocity. These are used to put into the slope formula (both 0 aren't needed):
30 / 10 = 3 m/s2
Problem 2
Problem 2
When a hockey player hits a puck with his stick, the puck's velocity changes from 8.0 m/s [N] to 10.0 m/s [S] in 0.05 seconds.
What's the puck's acceleration?
Solution:
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aav = 8m/s [N] - 10m/s [S] / 0.05 s (the vector subtraction can be changed into a vector addition, as two opposite
directions can have their subtraction reversed: North and South.)
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aav = 8m/s [N] + 10m/s [S] / 0.05 s
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aav = 360 m/s
See that the puck's initial velocity is north and its final is south. Its acceleration is from the initial motion, so the puck slows down and stops before accelerating south, increasing its velocity to 10 m/s [S], which is why the final velocity is due south
To calculate the displacement, it should be separated into a rectangle and a triangle and readd them together after calculating each.
Problem 3
Problem 3
What's the displacement shown in Figure 4's graph over the time interval from 0 s to 10.0 s?
Figure 4 velocity-time graph is more complex than previous ones. But the object's displacement can still be found by calculating the area under the velocity time graph.
Solution:
Triangle area: 1/2 x base x height
1/2 x 5x10 [S] = 25m2 [S]
Rectangle area: 10x5 = 50m2 [S]
Add together: 25[S] + 50[S] = 75m2[S]
The object has travelled 75 m [S] after 10.0 s.
Instantaneous velocity and average velocity
Instantaneous velocity and average velocity
Any object's velocity that's accelerating is changing overtime. In motion with uniform acceleration, an object's velocity changes at constant (uniform) rate. In a launch, a spacecraft accelerates upward at a rapid rate. NASA plots position and time data on a graph and determine their spacecraft's instantaneous velocity to to determine its velocity at specific points in time.
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Instantaneous velocity, or v ins , is an object'velocity at a specific instant in time. By comparing,
Instantaneous velocity (V inst): An object's velocity at a specific instant in time
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average velocity (v av) is determined overtime interval.
Both types of velocity are rates of changes of position, but they tell us various things for an
object's motion.
A plane mirror can be used to draw a tangent to a curved line. Place the mirror as perpendicular as possible to the line at the point desired.
Problem 4
Problem 4
Figure 6 shows an object's position-time graph undergoing uniform acceleration. Moving along the curve, the curve's slope progressively increases.
From this, it shows that the velocity of the object constantly increases.
To find the instantaneous velocity of the object at a specific time, the slope of the tangent of the line in the graph at that time must be calculated.
Adjust the mirror's angle so that the real curve merges with its images in the mirror, which occurs when the mirror is perpendicular to the curved line at the point.
Draw a line perpendicular to the mirror to obtain the curve's tangent.
Figure 6 can be help finding an object's instantaneous velocity at any point in time.
To find instantaneous velocity = tangent of line's slope must be determined
Determining the instantaneous velocity
Determining the instantaneous velocity
Consider the curve's point at 2.0 s on the x-axis.
What's the object's instantaneous at this time?
Solution:
Get 2 point on the tangent to find the slope (instantaneous velocity):
(2, 4) and (3, 8)
→
v ins = (8-4)/(3-2) = 4 [E]
The object's instantaneous velocity at 2.0 s is 4.0 m/s [E].
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