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Sagot :
To prove that the formula [tex]\( s = \frac{u+v}{2} \times t \)[/tex] is correct, let's start from fundamental principles of motion under uniform acceleration.
1. Given:
- [tex]\( u \)[/tex] is the initial velocity.
- [tex]\( v \)[/tex] is the final velocity.
- [tex]\( t \)[/tex] is the time taken.
- [tex]\( s \)[/tex] is the distance traveled.
2. Distance formula under uniform acceleration:
The uniformly accelerated motion can be described by the equation:
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
Here, [tex]\( a \)[/tex] is the acceleration.
3. Final velocity relation:
The final velocity [tex]\( v \)[/tex] can be expressed in terms of the initial velocity and acceleration:
[tex]\[ v = u + at \][/tex]
From this equation, we can solve for acceleration [tex]\( a \)[/tex]:
[tex]\[ a = \frac{v - u}{t} \][/tex]
4. Substitute [tex]\( a \)[/tex] in the distance formula:
Now, substitute [tex]\( a = \frac{v - u}{t} \)[/tex] into the distance formula [tex]\( s = ut + \frac{1}{2}at^2 \)[/tex]:
[tex]\[ s = ut + \frac{1}{2} \left(\frac{v - u}{t}\right) t^2 \][/tex]
5. Simplify the equation:
Simplify the expression:
[tex]\[ s = ut + \frac{1}{2} (v - u) t \][/tex]
Distribute the [tex]\( t \)[/tex] in the second term:
[tex]\[ s = ut + \frac{1}{2} v t - \frac{1}{2} u t \][/tex]
Combine the like terms:
[tex]\[ s = ut - \frac{1}{2} u t + \frac{1}{2} v t \][/tex]
[tex]\[ s = \frac{1}{2} u t + \frac{1}{2} v t \][/tex]
Factor out [tex]\( t/2 \)[/tex]:
[tex]\[ s = \frac{t}{2} (u + v) \][/tex]
6. Rearrange the equation:
Finally, express this in a more familiar form:
[tex]\[ s = \frac{u + v}{2} \times t \][/tex]
Thus, we have shown that the formula [tex]\( s = \frac{u+v}{2} \times t \)[/tex] is correct. This completes the proof.
1. Given:
- [tex]\( u \)[/tex] is the initial velocity.
- [tex]\( v \)[/tex] is the final velocity.
- [tex]\( t \)[/tex] is the time taken.
- [tex]\( s \)[/tex] is the distance traveled.
2. Distance formula under uniform acceleration:
The uniformly accelerated motion can be described by the equation:
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
Here, [tex]\( a \)[/tex] is the acceleration.
3. Final velocity relation:
The final velocity [tex]\( v \)[/tex] can be expressed in terms of the initial velocity and acceleration:
[tex]\[ v = u + at \][/tex]
From this equation, we can solve for acceleration [tex]\( a \)[/tex]:
[tex]\[ a = \frac{v - u}{t} \][/tex]
4. Substitute [tex]\( a \)[/tex] in the distance formula:
Now, substitute [tex]\( a = \frac{v - u}{t} \)[/tex] into the distance formula [tex]\( s = ut + \frac{1}{2}at^2 \)[/tex]:
[tex]\[ s = ut + \frac{1}{2} \left(\frac{v - u}{t}\right) t^2 \][/tex]
5. Simplify the equation:
Simplify the expression:
[tex]\[ s = ut + \frac{1}{2} (v - u) t \][/tex]
Distribute the [tex]\( t \)[/tex] in the second term:
[tex]\[ s = ut + \frac{1}{2} v t - \frac{1}{2} u t \][/tex]
Combine the like terms:
[tex]\[ s = ut - \frac{1}{2} u t + \frac{1}{2} v t \][/tex]
[tex]\[ s = \frac{1}{2} u t + \frac{1}{2} v t \][/tex]
Factor out [tex]\( t/2 \)[/tex]:
[tex]\[ s = \frac{t}{2} (u + v) \][/tex]
6. Rearrange the equation:
Finally, express this in a more familiar form:
[tex]\[ s = \frac{u + v}{2} \times t \][/tex]
Thus, we have shown that the formula [tex]\( s = \frac{u+v}{2} \times t \)[/tex] is correct. This completes the proof.
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