Physics, 22.09.2019 19:30 bella51032

Mary and sally are in a foot race. when mary is 22 m from the finish line, she has a speed of 4.0 m/s and is 5.0 m behind sally, who has a speed of 5.0 m/s. sally thinks she has an easy win and so, during the remaining portion of the race, decelerates at a constant rate of 0.50 m/s2 to the finish line.

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Answer from: davidtrott5761

$0.604 m/s^{2}$

Explanation:

The complete question is:

Mary and Sally are in a foot race (see the figure). When Mary is 22 m from the finish line, she has a speed of4.0 m/s and is 5.0 m behind Sally, who has a speed of5.0 m/s. Sally thinks she has an easy win and so, during the remaining portion of the race, decelerates at a constant rate of 0.44m/s2 to the finish line.

What constant acceleration does Mary now need during the remaining portion of the race, if she wishes to cross the finish line side-by-side with Sally?

We have the following data:

Mary's speed: $V_{M}=4m/s$

Mary's distance to the finish line: $d_{M}=22 m$

Sally's speed: $V_{S}=5m/s$

Sally's distance to the finish line: $d_{S}=22 m- 5 m=17 m$

Sally's acceleration: $a_{S}=-0.44 m/s^{2}$ because we are told she decelerates

We need to find Mary's acceleration: $a_{M}$

But firstly, we have to find the time it takes Sally to reach the finish line, which is the same time Mary will have to reach the finish line. This can be done by the following equation:

$d_{S}=V_{S}t+\frac{1}{2}a_{S}t^{2}$ (1)

Rewritting the equation:

$\frac{1}{2}a_{S}t^{2}+V_{S}t-d_{S}=0$ (2)

$\frac{1}{2}(-0.44 m/s^{2})t^{2}+5m/s t-17m=0$ (3)

Multiplying by 1 each side of the equation:

$(0.22 m/s^{2})t^{2}-5m/s t+17m=0$ (4)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (5)

Where:

a=0.22

b=-5

c=17

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(0.22)(17)}}{2(0.22)}$ (6)

Solving (6) we take the smaller result (because the longer result is equivalent to the time it takes to Sally to reach the finish line, change her direction and cross the finish line again):

t=4.18 s (7) Now that we have the time, we can calulate Mary's acceleration

$d_{M}=V_{M}t+\frac{1}{2}a_{M}t^{2}$ (8)

Isolating $a_{M}$ :

$a_{M}=\frac{2(d_{M}-V_{M}t)}{t^{2}}$ (9)

$a_{M}=\frac{2(22m-(4 m/s)(4.18 s))}{(4.18 s)^{2}}$ (10)

Finally:

$a_{M}=0.604 m/s^{2}$ (11) This is the acceleration Mary needs if she wishes to cross the finish line side-by-side with Sally.

Mary and sally are in a foot race. when mary is 22 m from the finish line, she has a speed of 4.0 m/

Answer from: Quest

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

Answer from: Quest

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6 $\frac{m}{s^{2} }$

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

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Mary and sally are in a foot race. when mary is 22 m from the finish line, she has a speed of 4.0 m/...

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