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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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• Why do you need the area of the square? Since we are measuring the distance covered by the vehicle between the time that it was at velocities of 5 m/s and 13 m/s, don't we just use the triangle?
So wouldn't it be s=(v(f)-v(i))times time which is s=(8)4=32? •   No. Displacement is area under the velocity-time graph.
If you didn't count the square, you'd go 1/2*4*8 = 16 m (Area of a triangle).
If you counted the square only, it would show you how much displament you had covered if you go at a constant velocity of 5 m/s. This is (as shown) 20 m.

But in the first case you're accelerating right? So your velocity is going to increase, i.e its going to be more than 5m/s. Thus it is impossible for you to go only 16m, which is less than 20m(and not 32 as you had calculated).
The total displacement is 20m + 16m = 36m.
• 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
• What is an increasing acceleration called (i.e. ∆a/∆t)? • 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? • 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. • 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).   