Answer:
0.68 s
Explanation:
We are given that
Initial velocity of box=[tex]u=13m/s[/tex]
Final velocity of box=v=11.5 m/s
Distance=d=8.5 m
We have to find the time taken by box to slow by this amount.
We know that
[tex]v^2-u^2=2as[/tex]
Substitute the values
[tex](11.5)^2-(13)^2=2a(8.5)[/tex]
[tex]132.25-169=17a[/tex]
[tex]-36.75=17a[/tex]
[tex]a=\frac{-36.75}{17}=-2.2m/s^2[/tex]
We know that
Acceleration=[tex]a=\frac{v-u}{t}[/tex]
Substitute the values
[tex]-2.2=\frac{11.5-13}{t}[/tex]
[tex]-2.2=\frac{-1.5}{t}[/tex]
[tex]t=\frac{1.5}{2.2}=0.68 s[/tex]
Hence, the time taken by box to slow by this amount=0.68 s
"0.68 s" would be taken by the box to slow by this amount.
Given values are:
Initial velocity,
Final velocity,
Distance,
As we know,
→ [tex]v^2-u^2 = 2as[/tex]
By substituting the values, we get
→ [tex](11.5)^2-(13)^2 = 2a(8.5)[/tex]
→ [tex]132.25-169=17 a[/tex]
→ [tex]a = -\frac{36.75}{17}[/tex]
[tex]= -2.2 \ m/s^2[/tex]
hence,
→ Acceleration, [tex]a = \frac{v-u}{t}[/tex]
[tex]-2.2 = \frac{11.5-13}{t}[/tex]
[tex]t = \frac{1.5}{2.2}[/tex]
[tex]= 0.68 \ s[/tex]
Thus the above response is right.
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