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3 years ago

A shopping cart rolls from x=19.9m to x=5.40m with an average velocity of -0.418m/s. How much time did it take? (unit=s)

Physics

1 answer:

andrew11 [14]3 years ago

8 0

Given

initial position = Xi= 19.9m

Final position Xf = 5.4m

Average velocity= Va = -0.418m/s

it shows displacement is reverse.

To find t=?

As Va = (Xf- Xi) / t

t = (Xf-Xi) / ( Va)

t = ( 5.4-19.9) / (-0.418)

t = (-14.5 ) / (-0.418) (-ve sign cancel out at numerator and denominator)

t =34.69 s

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Answer:

(a) The collision lasts for 0.053 s.

(b) The deceleration is 180.5 m/s².

Explanation:

Given:

Initial velocity of the player (u) = 9.50 m/s

Final velocity of the player (v) = 0 m/s (Comes to a stop)

Displacement of the player (S) = 0.250 m

We know that, using equation of motion relating displacement (S), acceleration (a), initial velocity (u) and final velocity (v), we have:

$v^2=u^2+2aS$

Expressing in terms of 'a', we get:

$a=\frac{v^2-u^2}{2S}$

Plug in the given values and solve for 'a'. This gives,

$a=\frac{0-9.50^2}{2\times 0.250}\\\\a=\frac{-90.25}{0.5}=-180.5\ m/s^2$

Therefore, the acceleration of the player is -180.5 m/s². So, the deceleration is 180.5 m/s².

Now, using the first equation of motion, we have:

$v=u+at\\\\t=\frac{v-u}{a}$

Plug in the given values and solve for 't'. This gives,

$t=\frac{0-9.5}{-180.5}\\\\t=0.053\ s$

Therefore, the the collision will last for 0.053 s.

(a) The collision lasts for 0.053 s.

(b) The deceleration is 180.5 m/s².

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