Certainly! Let's solve this problem step-by-step.
### Case 1: Starting from a stationary position to [tex]\( 5 \, m/s \)[/tex] in [tex]\( 20 \, s \)[/tex]
1. Initial velocity ([tex]\( u_1 \)[/tex]): [tex]\( 0 \, m/s \)[/tex]
2. Final velocity ([tex]\( v_1 \)[/tex]): [tex]\( 5 \, m/s \)[/tex]
3. Time taken ([tex]\( t_1 \)[/tex]): [tex]\( 20 \, s \)[/tex]
To calculate the acceleration ([tex]\( a_1 \)[/tex]), we use the formula:
[tex]\[ a = \frac{v - u}{t} \][/tex]
Substituting the values:
[tex]\[ a_1 = \frac{5 \, m/s - 0 \, m/s}{20 \, s} \][/tex]
[tex]\[ a_1 = \frac{5}{20} \][/tex]
[tex]\[ a_1 = 0.25 \, m/s^2 \][/tex]
So, the acceleration in the first case is [tex]\( 0.25 \, m/s^2 \)[/tex].
### Case 2: From [tex]\( 5 \, m/s \)[/tex] to [tex]\( 3 \, m/s \)[/tex] in [tex]\( 6 \, s \)[/tex]
1. Initial velocity ([tex]\( u_2 \)[/tex]): [tex]\( 5 \, m/s \)[/tex]
2. Final velocity ([tex]\( v_2 \)[/tex]): [tex]\( 3 \, m/s \)[/tex]
3. Time taken ([tex]\( t_2 \)[/tex]): [tex]\( 6 \, s \)[/tex]
To calculate the acceleration ([tex]\( a_2 \)[/tex]), we again use the formula:
[tex]\[ a = \frac{v - u}{t} \][/tex]
Substituting the values:
[tex]\[ a_2 = \frac{3 \, m/s - 5 \, m/s}{6 \, s} \][/tex]
[tex]\[ a_2 = \frac{-2}{6} \][/tex]
[tex]\[ a_2 = -0.333 \, m/s^2 \][/tex]
So, the acceleration in the second case is [tex]\( -0.333 \, m/s^2 \)[/tex] (negative acceleration indicates deceleration).
### Summary
- Acceleration from stationary to [tex]\( 5 \, m/s \)[/tex] in [tex]\( 20 \, s \)[/tex]: [tex]\( 0.25 \, m/s^2 \)[/tex]
- Acceleration from [tex]\( 5 \, m/s \)[/tex] to [tex]\( 3 \, m/s \)[/tex] in [tex]\( 6 \, s \)[/tex]: [tex]\( -0.333 \, m/s^2 \)[/tex]
These are the accelerations for both cases.
