Answer :
Sure, let's derive the equations of motion for constant acceleration using the graphical method. We will use velocity-time graphs to help visualize and derive each equation.
### (i) [tex]\( v = u + at \)[/tex]
1. Imagine a velocity-time graph. The initial velocity [tex]\(u\)[/tex] is the velocity at time [tex]\(t=0\)[/tex]. This is represented by a point on the vertical axis (y-axis) of the velocity-time graph.
2. The velocity increases uniformly over time [tex]\(t\)[/tex] due to constant acceleration [tex]\(a\)[/tex]. Thus, the graph of velocity vs. time is a straight line with slope [tex]\(a\)[/tex].
3. The final velocity [tex]\(v\)[/tex] at time [tex]\(t\)[/tex] can be found from this straight line. The general equation of a straight line is given by:
[tex]\[ v = u + at \][/tex]
### (ii) [tex]\( s = ut + \frac{1}{2}at^2 \)[/tex]
1. Again, consider the velocity-time graph with initial velocity [tex]\(u\)[/tex].
2. The area under the velocity-time graph represents the displacement [tex]\(s\)[/tex].
3. For constant acceleration, the graph is a straight line with slope [tex]\(a\)[/tex]. Therefore, the area under the graph is a trapezoid.
4. The area of a trapezoid can be found by dividing it into two simpler shapes: a rectangle and a triangle.
- The rectangle's area, which extends from [tex]\(0\)[/tex] to [tex]\(t\)[/tex] (time) and up to [tex]\(u\)[/tex] (initial velocity), is [tex]\(u \times t\)[/tex].
- The triangle's area, which extends from [tex]\(u\)[/tex] to [tex]\(v\)[/tex] (final velocity) with the base as [tex]\(t\)[/tex] (time), is [tex]\(\frac{1}{2} \times t \times (v - u)\)[/tex].
5. Substituting [tex]\(v = u + at\)[/tex] from the first equation into the expression for the triangular area:
[tex]\[ \text{Area of triangle} = \frac{1}{2} \times t \times (u + at - u) = \frac{1}{2} \times t \times at = \frac{1}{2}at^2 \][/tex]
6. Therefore, the total displacement [tex]\(s\)[/tex] is:
[tex]\[ s = \text{Area of rectangle} + \text{Area of triangle} = ut + \frac{1}{2}at^2 \][/tex]
### (iii) [tex]\( v^2 = u^2 + 2as \)[/tex]
1. Using the first equation: [tex]\( v = u + at \)[/tex], solve for [tex]\(t\)[/tex]:
[tex]\[ t = \frac{v - u}{a} \][/tex]
2. Substitute this expression for [tex]\(t\)[/tex] into the second equation: [tex]\( s = ut + \frac{1}{2}at^2 \)[/tex]:
[tex]\[ s = u \left(\frac{v-u}{a}\right) + \frac{1}{2}a\left(\frac{v-u}{a}\right)^2 \][/tex]
3. Simplify the equation:
[tex]\[ s = \frac{u(v - u)}{a} + \frac{1}{2}a \cdot \frac{(v - u)^2}{a^2} \][/tex]
[tex]\[ s = \frac{uv - u^2}{a} + \frac{1}{2} \cdot \frac{(v - u)^2}{a} \][/tex]
[tex]\[ s = \frac{uv - u^2 + \frac{1}{2}(v^2 - 2uv + u^2)}{a} \][/tex]
[tex]\[ s = \frac{uv - u^2 + \frac{1}{2}v^2 - uv + \frac{1}{2}u^2}{a} \][/tex]
[tex]\[ s = \frac{\frac{1}{2}v^2 - \frac{1}{2}u^2}{a} \][/tex]
[tex]\[ 2as = v^2 - u^2 \][/tex]
4. Rearrange to get the final form:
[tex]\[ v^2 = u^2 + 2as \][/tex]
And there we have the three equations of motion derived using the graphical method: [tex]\(v = u + at\)[/tex], [tex]\(s = ut + \frac{1}{2}at^2\)[/tex], and [tex]\(v^2 = u^2 + 2as\)[/tex].
### (i) [tex]\( v = u + at \)[/tex]
1. Imagine a velocity-time graph. The initial velocity [tex]\(u\)[/tex] is the velocity at time [tex]\(t=0\)[/tex]. This is represented by a point on the vertical axis (y-axis) of the velocity-time graph.
2. The velocity increases uniformly over time [tex]\(t\)[/tex] due to constant acceleration [tex]\(a\)[/tex]. Thus, the graph of velocity vs. time is a straight line with slope [tex]\(a\)[/tex].
3. The final velocity [tex]\(v\)[/tex] at time [tex]\(t\)[/tex] can be found from this straight line. The general equation of a straight line is given by:
[tex]\[ v = u + at \][/tex]
### (ii) [tex]\( s = ut + \frac{1}{2}at^2 \)[/tex]
1. Again, consider the velocity-time graph with initial velocity [tex]\(u\)[/tex].
2. The area under the velocity-time graph represents the displacement [tex]\(s\)[/tex].
3. For constant acceleration, the graph is a straight line with slope [tex]\(a\)[/tex]. Therefore, the area under the graph is a trapezoid.
4. The area of a trapezoid can be found by dividing it into two simpler shapes: a rectangle and a triangle.
- The rectangle's area, which extends from [tex]\(0\)[/tex] to [tex]\(t\)[/tex] (time) and up to [tex]\(u\)[/tex] (initial velocity), is [tex]\(u \times t\)[/tex].
- The triangle's area, which extends from [tex]\(u\)[/tex] to [tex]\(v\)[/tex] (final velocity) with the base as [tex]\(t\)[/tex] (time), is [tex]\(\frac{1}{2} \times t \times (v - u)\)[/tex].
5. Substituting [tex]\(v = u + at\)[/tex] from the first equation into the expression for the triangular area:
[tex]\[ \text{Area of triangle} = \frac{1}{2} \times t \times (u + at - u) = \frac{1}{2} \times t \times at = \frac{1}{2}at^2 \][/tex]
6. Therefore, the total displacement [tex]\(s\)[/tex] is:
[tex]\[ s = \text{Area of rectangle} + \text{Area of triangle} = ut + \frac{1}{2}at^2 \][/tex]
### (iii) [tex]\( v^2 = u^2 + 2as \)[/tex]
1. Using the first equation: [tex]\( v = u + at \)[/tex], solve for [tex]\(t\)[/tex]:
[tex]\[ t = \frac{v - u}{a} \][/tex]
2. Substitute this expression for [tex]\(t\)[/tex] into the second equation: [tex]\( s = ut + \frac{1}{2}at^2 \)[/tex]:
[tex]\[ s = u \left(\frac{v-u}{a}\right) + \frac{1}{2}a\left(\frac{v-u}{a}\right)^2 \][/tex]
3. Simplify the equation:
[tex]\[ s = \frac{u(v - u)}{a} + \frac{1}{2}a \cdot \frac{(v - u)^2}{a^2} \][/tex]
[tex]\[ s = \frac{uv - u^2}{a} + \frac{1}{2} \cdot \frac{(v - u)^2}{a} \][/tex]
[tex]\[ s = \frac{uv - u^2 + \frac{1}{2}(v^2 - 2uv + u^2)}{a} \][/tex]
[tex]\[ s = \frac{uv - u^2 + \frac{1}{2}v^2 - uv + \frac{1}{2}u^2}{a} \][/tex]
[tex]\[ s = \frac{\frac{1}{2}v^2 - \frac{1}{2}u^2}{a} \][/tex]
[tex]\[ 2as = v^2 - u^2 \][/tex]
4. Rearrange to get the final form:
[tex]\[ v^2 = u^2 + 2as \][/tex]
And there we have the three equations of motion derived using the graphical method: [tex]\(v = u + at\)[/tex], [tex]\(s = ut + \frac{1}{2}at^2\)[/tex], and [tex]\(v^2 = u^2 + 2as\)[/tex].
