Question

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?"

Answer 1

Answer:

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