Hello there. Last week in class we learned five new equations called the Big 5, each interlocked with one another. Here are the five equations we learned:
1) The DAD at=V2-V1
2) The MOM d= ½(V1+V2)Δt
3) First CHILD d=V1t + ½at²
4) Second CHILD d=V2t - ½at²
5) Third CHILD V2²= V1²+2ad
From these five equations, you must also be able to derive them from a graph. In order to get equation 3, you must find the area of the small triangle from the graph and the area of the rectangle and add them together.
Area of a triangle: d= 1/2 (V2 -V1 )t
Substitute (at) for (V2-V1) of equation 2.
d= 1/2 (at)t + V1 t
d= 1/2 a t2 + V1 t
d= V1 t + 1/2 a t2
In order to get equation 4, you must subtract the area of the triangle on top, from the large rectangle.
d= V2 t - 1/2 (V2 -V1 )t
Substitute (at) for (V2 -V1 ) of equation 2
d= V2 t - 1/2 (at )t
d= V2 t - 1/2 a t2
Once you have fully understood these methods, you will be ready to tackle future tasks that come your way!
Thanks,
-Peggy.
1) The DAD at=V2-V1
2) The MOM d= ½(V1+V2)Δt
3) First CHILD d=V1t + ½at²
4) Second CHILD d=V2t - ½at²
5) Third CHILD V2²= V1²+2ad
From these five equations, you must also be able to derive them from a graph. In order to get equation 3, you must find the area of the small triangle from the graph and the area of the rectangle and add them together.
Area of a triangle: d= 1/2 (V2 -V1 )t
Substitute (at) for (V2-V1) of equation 2.
d= 1/2 (at)t + V1 t
d= 1/2 a t2 + V1 t
d= V1 t + 1/2 a t2
In order to get equation 4, you must subtract the area of the triangle on top, from the large rectangle.
d= V2 t - 1/2 (V2 -V1 )t
Substitute (at) for (V2 -V1 ) of equation 2
d= V2 t - 1/2 (at )t
d= V2 t - 1/2 a t2
Once you have fully understood these methods, you will be ready to tackle future tasks that come your way!
Thanks,
-Peggy.
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