Physics, 12.02.2020 19:20 yeontan

The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5 m. "How long in seconds did it take for the box to slow by this amount, assuming the acceleration is constant?"

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Answer from: elizaannettem4466

"0.68 s" would be taken by the box to slow by this amount.

Given values are:

Initial velocity,

u = 13 m/s

Final velocity,

v = 11.5 m/s

Distance,

d = 8.5 m

As we know,

→ v^2-u^2 = 2as

By substituting the values, we get

→ (11.5)^2-(13)^2 = 2a(8.5)

→ 132.25-169=17 a

→ $a = -\frac{36.75}{17}$

$= -2.2 \ m/s^2$

hence,

→ Acceleration, $a = \frac{v-u}{t}$

$-2.2 = \frac{11.5-13}{t}$

$t = \frac{1.5}{2.2}$

$= 0.68 \ s$

Thus the above response is right.

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Answer from: zanestone12

0.68 s

Explanation:

We are given that

Initial velocity of box= u=13m/s

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

v^2-u^2=2as

Substitute the values

(11.5)^2-(13)^2=2a(8.5)

132.25-169=17a

-36.75=17a

$a=\frac{-36.75}{17}=-2.2m/s^2$

We know that

Acceleration= $a=\frac{v-u}{t}$

Substitute the values

$-2.2=\frac{11.5-13}{t}$

$-2.2=\frac{-1.5}{t}$

$t=\frac{1.5}{2.2}=0.68 s$

Hence, the time taken by box to slow by this amount=0.68 s

Answer from: Quest

come closer and become far apart and become steam/gas

Answer from: Quest

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The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5...

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