Answer:
The answer is "91 m"
Explanation:
[tex]v_0 = 2.5 \ \frac{m}{s}\\\\a = 4.2\ \frac{m}{s^2}\\\\t = 6.0\ s\\\\\Delta x=?[/tex]
Using formula:
[tex]\Delta x = v_0 t +\frac{1}{2} at^2\\\\[/tex]
[tex]= 2.5 \times 6.0 +\frac{1}{2} \times 4.2 \times 6.0^2\\\\= 90.6\ m \approx 91\ m[/tex]
