Once the child in the sample problem reaches the bottom of the hill, she continues sliding along the flat, snow-covered ground until she comes to a stop. If her acceleration during this time is -.392m/s^2 how long does it take her to travel from the bottom of the hill to her stopping point

Sample problem is;

A child slides down a steep, snow covered hill with an acceleration of 2.82 m/s². Her initial speed is 0.0 m/s² and her final speed is 15.5 m/s travelling from top to bottom of the hill.

t = 39.54 s

Step-by-step explanation:

We are told that her acceleration is now -0.392 m/s²

Thus; a = -0.392 m/s²

From the sample problem, we have;. Initial velocity = 0 m/s

Final velocity = 15.5 m/s

However, in this question, we want to find the time from the bottom of the hill to her stopping point.

This means her initial velocity will now be; u = 15.5 m/s and final velocity since she will come to rest will be v = 0 m/s

To find the time it takes her to travel from the bottom of the hill to her stopping point, we will use Newton's equation of motion and so, we have;

v = u + at

making t the subject, we have;

t = (v - u)/a

t = (0 - 15.5)/(-0.392)

t = 39.54 s

529 and 531

step-by-step explanation:

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°