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Answer:
1. Acceleration:
[tex]{\( = 5\ m/s^2}[/tex]
2. Nearest Average Speed: within options
[tex]\(= 3\ m/s\ approx\).[/tex]
Explanation:
[tex]- \( s = 4 \) meters at \( t = 3 \) seconds.[/tex]
Using the formula for displacement under constant acceleration:
[tex]\[ s = ut + \frac{1}{2} a t^2 \][/tex]
Assuming the car starts from rest ( u = 0):
[tex]4 = 0 \cdot 3 + \frac{1}{2} a (3^2)[/tex]
[tex]4 = \frac{1}{2} a \cdot 9[/tex]
[tex]a = \frac{8}{9} \approx 0.89\ \text{m/s}^2[/tex]
The above method gives us acceleration for a time period but 2-5 dynamic checks may alter:
### Checking options (iterative):
1. Given larger options practical:
[tex]a = 5\ m/s^2 [/tex]
[tex]{- Distance \( s = 4 \) meters at \( t = 3 \) \: seconds.}[/tex]
[tex]{\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{4}{3}\ \text{m/s} \approx 1.33\ \text{m/s}\]}[/tex]