Everything Science Grade 10

CHAPTER 21. MOTION IN ONE DIMENSION 21.7

~a=∆t~v

where∆~vis the change in velocity, i.e.∆v=~vf-~vi. Thus we have

~a = ~vf−t~vi

~vf = ~vi+~at

Derivation of 21.2

We have seen that displacement can be calculated from the area under a velocity vs. time
graph. Foruniformly accelerated motionthe most complicated velocity vs. time graph we
can have is a straight line. Look at the graph below - it represents an object with a starting
velocity of~vi, accelerating to a final velocity~vfover a total timet.

~v(m·s−^1 )

t(s)

~vi

~vf

t

To calculate the final displacement we must calculate the area under the graph - this is just
the area of the rectangle added to the area of the triangle. This portion of the graph has
been shaded for clarity.

Area△ =^12 b×h

=^12 t×(vf−vi)

=^12 vft−^12 vit

Area = ℓ×b

= t×vi

= vit

Physics: Mechanics 431