# Math Physics Essay

QUESTIONS 1: While following a treasure map, you start at an old oak tree. You first walk 825m directly south, then turn and walk 1.25km at 30 degrees west of north, and finally walk 1.00km at 40 degrees north of east,where you find the treasure: a biography of Issac Newton! (a) To return to the old oak tree, in what direction should you head and how far will you walk? USE COMPONENTS TO SOLVE THIS PROBLEM.

To solve the problem, we can compute for the resultant displacement — the net distance travelled, and its direction relative to the OAK tree.

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The promptest way to go back to the OAK tree is to walk towards the opposite direction of the resultant displacement.

The component method involves breaking down the vector into x and y components. The resultant vector is determined by adding all x components and the y-components.

Displacement X-component y-component

———— ———– ———–

825 m South 0 -825

1,250 m at 300 W of N -1,250 sin(300) 1,250 cos(300)

1,000 m at 400 N of E 1,000 cos(400) 1,000 sin(400)

———– ———–

Resultant Displacement -1,391.04 900.32

The person has actually travelled 900 meters towards the north, and 1,391 meters towards the west.

To determine its magnitude, we will use the Pythagorean Theorem.

RD = SQRT[(-1,391.04)2 + (900.32)2] = 1,657 meters.

Since the x-component is negative, and the y-component is positive, the resultant displacement is on the second quadrant.

To determine its direction, we note that the:

tangent(angle) = 900.32/-1,391.04

angle = tan-1 (900.32/-1,391.04)

= -570

The person, in all has walked 1,657 meters, about 570 north of west from the oak tree towards the treasure map. In other words, the treasure map is located 1,657 meters, about 57% north of west from the oak tree.

Hence, to go back to the oak tree from the treasure map, the same person will just walk 1,657 meters and walk 57% south of east.

QUESTION 2: According to recent test data, an automobile travels 0.025 miles in 19.9s, starting from rest. The same car, when braking from 60.0miles per hour on dry pavement, stops in 146ft. Assume constant acceleration when slowing down as when speeding up. (a) Find the acceleration of this car when it is speeding up and when it is braking. (b) If its acceleration is constant, how fast (in miles per hour) should this car be traveling after 0.025 miles of acceleration? The actual measured speed is 70.0mi/h; what does this tell you about the motion? How long does it take this car to stop while braking from 60.0 mi/h?

(a) acceleration.

These formulas are applied in basic mechanics

vf= vo + at; or a = (vf – vo)/t,

x = vo t + 0.5at2; or a = 2[x – vo t]/t2

vf2 = vo 2 + 2ax; or a = (vf2 – vo 2)/(2x)

where:

vf is final velocity velocity

vo is initial velocity

a is acceleration

t is time

x is distance travelled

(a)(1) acceleration when it is speeding up?

X = 0.025 miles

Time = 19.9 seconds

Vo = starting from rest = 0 miles per second

Use the applicable formula,

a = 2[x – vo t]/t2

= 2(.025 miles – 0(19.9 second))/ (19.9 second)2

= .000126259 miles per second2

= .000126259 miles per second2 (5280 feet/ 1 mile)

= 6.613 feet per second2

The car accelerates at 6.613 feet per second2.

(a)(2) acceleration when it is braking?

Given:

Vo = 60.0 miles per hour

x = 146ft.

vf = stop = 0 miles per hour.

We need to have consistent units. Hence, we can change feet to miles.

Vo = 60.0 miles per hour

x = 146ft. (1 mile / 5280 feet) = .027652 miles.

vf = stop = 0 miles per hour.

Applicable formula is:

a = (vf2 – vo 2)/(2x)

= [(0 miles per hour)2 – (60 feet per hour) 2/ (2 * .027652 miles)

= – 65,096 miles per hour2

= – 65,096 miles per hour2 (1 hour/3600 seconds2)

(5,280 feet/ mile).

= – 26.52 feet per second

When the car is braking, it accelerates at -26.52 feet per second, or decelerates at 26.52 feet per second.

(b) If its acceleration is constant, how fast (in miles per hour) should this car be traveling after 0.025 miles of acceleration? The actual measured speed is 70.0mi/h; what does this tell you about the motion? How long does it take this car to stop while braking from 60.0 mi/h?

(b)(1) velocity?

vf2 = vo 2 + 2ax

vf = SQRT( vo 2 + 2ax) =

vf = SQRT((0)2 + 2(.000126259 miles per second2)(0.025 miles)

= .002513 miles per second

= 9.045 miles per hour

If the actual measured speed is 70.0 mi/h, it means that acceleration was not constant at all.

(b)(2) time to stop?

vf = vo + at; or

t =(vf – vo)/a

= (0 – 70.0 mi/hr.)/ (- 65,096 miles per hour2)

= 3.87 seconds

QUESTION 3: An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the way down and passes a point 50.0m below its starting point 5.00s after it leaves the throwers hand. Air resistance may be ignored. (a) What is the initial speed of the egg? (b) how high does it riseabove the starting point? (c) What is the magnitude of the velocity at the highest point? (d) What are the magnitude and direction of its acceleration at the highest point?

(a) What is initial speed of the egg?

t = 5 seconds

x = – 50 meters

g = -9.8 m/s2

Use this formula x = vo t + 0.5gt2

vo = (x – 0.5gt2)/t

= (-50 – 0.5 (-9.8)(5)2)/5

= 14.5 meters / second

(b) how high does it rise above the starting point?

vf2 = vo2 + 2ax

x = (vf2 – vo2)/2g

The egg reaches its maximum point when velocity is zero. Hence,

x = (0 – 14.52)/[2(-9.8)

= 10.73 meters.

(c) What is the magnitude of the velocity at the highest point?

The magnitude of the velocity at the highest point is zero.

(d) What are the magnitude and direction of its acceleration at the highest point?

Acceleration at the highest point is -9.8 m/s2

QUESTION 4: A 76.0-kg boulder is rolling horizontally at the top of a vertical cliff that is 20 m above the surface of a lake. The top of the vertical face of a dam is located 100m from the foot of the cliff, with the top of the dam level with the surface of the water in the lake. A level plain is 25 m below the top of the dam. (a) what must be the minimum speed of the rock just as it leaves the cliss so it will travel to the plain without striking the dam? (b) how far from the foot of the dam does the rock hit the plain?

(a) what must be the minimum speed of the rock just as it leaves the cliff so it will travel to the plain without striking the dam?

Using components, the distance the boulder will travel in the horizontal will depend on its speed while it leaves the cliff. As for the vertical, the speed will depend on gravitational acceleration.

To reach the plain, the boulder must travel 20 meters downward and 100 meters in the horizontal.

Consider the vertical first:

x = vo t + 0.5at2

vo = 0

x = 20

a = -9.8

x = (0)t + 0.5at2

x= .5 at2

t = sqrt(2x/g)

t = 2.0203 seconds

Consider the horizontal

Since, the time to reach the tip of the dam should be consistent with the vertical

Speed = distance/time

Speed = 100 meters / 2.202 seconds

Speed = 50 meters per second.

Hence, the boulder should be rolling at least 50 meters per second in order to avoid hitting the dam.

(b) how far from the foot of the dam does the rock hit the plain?

To do that, determine the time the rock reaches the bottom.

x = (0)t + 0.5at2

x= .5 at2

t = sqrt(2(45)/g)

t = 3.03 seconds

distance travelled on the horizontal is

distance = speed * time

distance = 50 m/s * 3.03 = 151.523 meters

Since the lake is 100 meters wide, the rock will travel 51.5 meters from the dam.