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# Average velocity for constant acceleration

Investigate the relationships between velocity, acceleration, and distance in a physics context. Learn how to use formulas to calculate final velocity and total distance traveled given an initial velocity, constant acceleration, and time. Created by Sal Khan.

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• What is an increasing acceleration called (i.e. ∆a/∆t)?
• yes !, jerk is also known as change in its inertial position of an object by any force. In jerk ( impulse) FORCE(F) is crucial. so, we knew that F = mass * acceleration. Thus
by this force is proportional to acceleration. so the above statement is justified.!
• in my physics book i have another formula of distance:
it's initial distance + intial velocity x time +1/2 acceleration x time^2
is this formula other way of writting the formula of sal?
• The formula you listed in the first question is a derivation of the sum of areas of the rectangle and triangle. Here S is displacement, u is initial velocity, v is final velocity, A is acceleration and t is time.
D = ut + (1/2)(v-u)t
(Multiplying and dividing t)
D = ut + (1/2)[(v-u)/t]t.t
[(v-u)/t is Acceleration]
D = ut + (1/2)A(t^2)
This can be also written as D = Initial Velocity x Time + 1/2 Acceleration x Time^2
• At why did sal use a triangle before the t?
• The ∆ means "change" and is frequently used in physics. I think the "t" stands for time, so what he was trying to say is "change in time" meaning how much time has passed for a certain process to complete or for a certain event to take place.
• At , I'm having trouble understanding why the equation can also be called as the average velocity. I remember from the previous videos that average velocity is the displacement over time. How can (1/2)(change in t)(v(i)+v(f)) be the average velocity?
• Sal does not include ∆t in the formula for average velocity. Velocity = Displacement/Time. Average Velocity only equals Displacement/Time when the Velocity is constant. When Velocity is not constant, Average Velocity = Initial Velocity + Final Velocity/2( the average of V(i) and V(f) )
• Am I correct in assuming the logic still holds by using the area of a trapezoid, instead of triangle plus square?
1/2 x (a+b) x h
So that'd be:
1/2 x (vi+vf) x t, which is what you got in the video.
• Yes, you can do it this way. Most of the students don't know if it's a trapezoid and do not remember its area. Sal thus chose a better way to find the area even if it was some other shape than a trapezoid. But in this case, you are assuming right
(1 vote)
• When Sal used 8m/s in the beginning and in triangle formula....Where did he get it from?
• Sal writes 8m/s for the first time at in the video. He started with the formula:
vi + (delta t)(a) = vf
5 + (4s)(2 m/s^2) = vf
5 m/s + 8 m/s = vf
13 m/s = vf
So 8 m/s = (delta t)(a) = vf - vi = delta v
This 8 m/s is the change in velocity (v increased by 2 m/s every second, for 4 seconds).
That's why 8 m/s also became the height of the triangle, since the bottom of the triangle on the graph was at vi = 5 m/s, and the top of the triangle was at vf = 13 m/s. The graph (velocity vs time) is a straight line that rises with a constant slope of 2, because the velocity is increasing at a steady rate of 2 m/s (graph goes up 2) per second (graph goes right 1).
• I'm new here, sorry if I ask a stupid question.

When Sal calculates the distance in the second half of the video. Wouldn't it be right to call it the displacement? My reasoning is that since he is calculating with velocity (a vector) and not just speed, the result wood be a vector again.

(I am aware that in the given example distance and displacemet are the same).
• Since, we are working with "constant acceleration" (magnitude and direction of acceleration will not change) motion will be in a straight line. So, we can use "displacement" and "distance" interchangeably.
• Starting from Sal breaks the integral of the velocity-time curve into two parts. But wouldn't it be easier if he used the formula A(trapezoid) = 1/2(side 1 + side 2)*height, which in this case would be 1/2(v1 + v2)*deltaT? One step versus two steps.
• Yes, but it is easier and more clear in his way, since in physics math is not the biggest challenge
• Do you also cancel units when you are multiplying? In this video, you cancel second (s) when you multiplied 4 seconds times 2 meters per second. I thought we only cancel units when we use division.
• You "cancel" units the same way you cancel numbers.
It's not really "cancelling". It's recognizing that n/n = 1.
• What does Sal do in where he says that v(i) and 1/2 v(i) will simplify to 1/2 v(i)?