Answer:
the time taken for the box to stop is 3.6 s.
Explanation:
Given;
initial speed with which the box was pushed, u = 5 m/s
time taken for the box to stop, t = 1.5 s
Assuming the box is moving at constant acceleration, a;
[tex]a = \frac{u}{t} \\\\a = \frac{5}{1.5} \\\\a = 3.33 \ m/s[/tex]
When the box is pushed a speed of 12 m/s, the time taken for the box to stop is calculated as;
[tex]t = \frac{v}{a} \\\\t = \frac{12}{3.33} \\\\t = 3.6 \ s[/tex]
Therefore, the time taken for the box to stop is 3.6 s.
