Viraam said:
Actually, this question is appearing in our Chapter called Force and motion where the following equations are given:
## F=ma \\ F= \frac{m(v-u)}{t} \\ v = u+at \\ v^2 - u^2 = 2as \\ s = ut + \frac{1}{2} at^2 \\ \text{Conservation of Linear Momentum} ##
So did they explain where these equations come from? I suppose they just presented them.
So you have been using: $$\sum_{net}F = F_1 + F_2 + ... + F_i = m \times a$$
Now acceleration (a) is the rate of of change in velocity with respect to time.: $$ a = \frac {\delta V} {\delta Time} = \frac {V_2 - V_1} {T_2 - T_1}$$
An object with zero initial velocity is accelerated at 10 m/s^2 for 10 seconds. What is the final velocity?
...$$ a = \frac {\delta V} {\delta Time} = 10 m/s^2= \frac {V_2 - 0} {10 - 0}$$
Rearrange to solve for V_2
...$$ V_2 = 10 m / s^2 * 10 s= 100 m/s$$
graphical solution:
On graph paper, Draw a graph of gravitational acceleration (use 10 m/s^2 for simpler math) versus time for time from zero to ten seconds. It is a nice flat line. The change in velocity is simply the shaded area under the curve ! So count off the number of boxes for each second.
In the first second, we have 10 boxes, so we are at 10 m/s after the first second.
The second second adds ten more boxes, so we are now at 20 m/s.
The third second adds ten more, so 30 m/s
4,5,6,7,8,9. Are 40, 50, 60, 70 , 80, and 90 m/s respectively.
At 10 seconds, we will have 100 squares, and thus be at 100 m/s.
On graph paper, draw the velocity as a function of time for ten seconds. Notice that it is a nice straight line with a slope of 10 m/s.
Integration:
When we integrate, we are just adding up the area under the curve!
Position:
Now velocity (v) is just the rate of change in position with respect to time. $$ v = \frac {\delta S} {\delta Time} = \frac {S_2 - S_1} {T_2 - T_1}$$
If the initial position was zero, what is the final position? We would need to use the average velocity (50 m/s is 1/2 for 100 and 0)
$$ S_2 = 50 m/s * 10 seconds + o m = 500 m$$
Graphical solution:
Count up the squares underneath the curve (straight line)
So for 0 seconds, 0 m
1 second... 0+ 10 = 10 m
2 seconds. ... 10 + 20 = 30 m
3 s...... 30 + 30 = 60 m
...
10 s ....,,,,, 400 + 100 = 500 m
Plot your result position versus time on your graph paper. You should have a nice curve.
Note: if we go the other way, position to velocity to acceleration, we reverse the process. We 'differentiate'. No you are finding the slope of the line at each instant in time.
The slope of a squared line (position) is a straight line.
The slope of a straight line is a constant.
