To derive the second equation of motion, we start with the first equation of motion, which relates the displacement (s) of an object to its initial velocity (u), acceleration (a), and time (t):
\[ s = ut + \frac{1}{2}at^2 \]
Now, let's derive the second equation of motion by eliminating displacement (s) from the first equation of motion. To do this, we'll solve the first equation for time (t):
\[ s = ut + \frac{1}{2}at^2 \]
Rearrange the equation:
\[ 2s = 2ut + at^2 \]
Now, let's isolate the term with t^2:
\[ at^2 = 2s - 2ut \]
Next, divide the entire equation by a:
[tex]\[ t^2 = \frac{2s - 2ut}{a} \][/tex]
Finally, take the square root of both sides to solve for time (t):
[tex]\[ t = \sqrt{\frac{2s - 2ut}{a}} \][/tex]
This is the derived second equation of motion. It relates the time (t) taken by an object to travel a distance (s) with initial velocity (u) and acceleration (a). Note that this equation applies only when the acceleration is constant.
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Answer:
s = ut + 1 / 2at²
[tex] \rule{200pt}{3pt}[/tex]
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Verified answer
To derive the second equation of motion, we start with the first equation of motion, which relates the displacement (s) of an object to its initial velocity (u), acceleration (a), and time (t):
\[ s = ut + \frac{1}{2}at^2 \]
Now, let's derive the second equation of motion by eliminating displacement (s) from the first equation of motion. To do this, we'll solve the first equation for time (t):
\[ s = ut + \frac{1}{2}at^2 \]
Rearrange the equation:
\[ 2s = 2ut + at^2 \]
Now, let's isolate the term with t^2:
\[ at^2 = 2s - 2ut \]
Next, divide the entire equation by a:
[tex]\[ t^2 = \frac{2s - 2ut}{a} \][/tex]
Finally, take the square root of both sides to solve for time (t):
[tex]\[ t = \sqrt{\frac{2s - 2ut}{a}} \][/tex]
This is the derived second equation of motion. It relates the time (t) taken by an object to travel a distance (s) with initial velocity (u) and acceleration (a). Note that this equation applies only when the acceleration is constant.