To solve this problem, we can use one of the kinematic equations that relate initial position, initial velocity, acceleration, time, and final position. The relevant equation in this context is:
[tex]\[ s = s_0 + v_0 t + \frac{1}{2} a t^2 \][/tex]
where:
- [tex]\( s \)[/tex] is the final position
- [tex]\( s_0 \)[/tex] is the initial position
- [tex]\( v_0 \)[/tex] is the initial velocity
- [tex]\( a \)[/tex] is the acceleration
- [tex]\( t \)[/tex] is the time
Given the following values:
- [tex]\( s_0 = -20 \)[/tex] meters (since the object starts 20 m to the left of the origin)
- [tex]\( v_0 = 10 \)[/tex] meters per second
- [tex]\( a = 4 \)[/tex] meters per second squared
- [tex]\( t = 5 \)[/tex] seconds
We can plug these values into the kinematic equation:
[tex]\[ s = -20 + (10 \times 5) + \frac{1}{2} (4 \times 5^2) \][/tex]
First, calculate the term with the initial velocity:
[tex]\[ 10 \times 5 = 50 \][/tex]
Next, calculate the term with the acceleration:
[tex]\[ \frac{1}{2} (4 \times 5^2) \][/tex]
[tex]\[ 5^2 = 25 \][/tex]
[tex]\[ 4 \times 25 = 100 \][/tex]
[tex]\[ \frac{1}{2} \times 100 = 50 \][/tex]
Now substitute these values back into the equation:
[tex]\[ s = -20 + 50 + 50 \][/tex]
Add these terms together:
[tex]\[ s = 80 \][/tex]
Thus, the position of the object from the origin at the end of the 5 seconds of acceleration is 80 meters to the right of the origin.
