4. AN OBJECT INCREASES ITS VELOCITY

A van starts off 152 miles directly north from the city of Springfield. It travels due east at a speed of 25 miles per hour. Aft

erastovalidia [21]

Answer:

12.84 miles per hour

Explanation:

Given:

Vertical distance of starting point of van from Springfield (d) = 152 miles

Speed in east direction (s) = 25 mph

Distance traveled in east direction (e) = 91 miles

Let the direct distance from Springfield of the van be 'x' at any time 't'.

Now, from the question, it is clear that, the vertical distance of van is fixed at 152 miles and only the horizontal distance is changing with time.

Now, consider a right angled triangle SNE representing the given situation.

Point S represents Springfield, N represents the starting point of van and E represents the position of van at any time 't'.

SN = d = 152 miles (fixed)

Now, using the pythagorean theorem, we have:

$SE^2=SN^2+NE^2\\\\x^2=d^2+e^2\\\\x^2=(152)^2+e^2----(1)$

Now, differentiating both sides with respect to time 't', we get:

$2x\frac{dx}{dt}=0+2e\frac{de}{dt}\\\\\frac{dx}{dt}=\frac{e}{x}\frac{de}{dt}$

Now, we are given speed as 25 mph. So, $\frac{de}{dt}=25\ mph$

Also, when $e=91\ mi$ , we can find 'x' using equation (1). This gives,

$x^2=23104+(91)^2\\\\x=\sqrt{31385}=177.16\ mi$

Now, plug in the values of 'e' and 'x' and solve for $\frac{dx}{dt}$ . This gives,

$\frac{dx}{dt}=\frac{91}{177.16}\times 25\\\\\frac{dx}{dt}=12.84\ mph$

Therefore, the distance between the van and Springfield is changing at a rate of 12.84 miles per hour

6 0

3 years ago

Calculate the acceleration of a car if it's velocity increases from 15 m/s to 75m/s in 5 seconds.

Andreas93 [3]

initial velocity=u=15m/s
Final velocity=v=75m/s
Time=t=5s

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{75-15}{5}$

$\\ \sf\longmapsto Acceleration=\dfrac{60}{5}$

$\\ \sf\longmapsto Acceleration=12m/s^2$

7 0

2 years ago

Help<br><br><br> What is Gravitational Potenial Energy

frosja888 [35]

Gravitational potential energy is an object that possesses because of its position in a gravitational field

5 0

2 years ago

What is the magnitude of g at a height above Earth's surface where free-fall acceleration equals 6.5m/s^2?

prohojiy [21]

You've given the answer, right there in your question.

The "magnitude of gravity" is described in terms of the acceleration
due to it, and you just told us what that is.

We can also notice that the figure you gave is about 0.66 of the
acceleration due to gravity on the Earth's surface. That tells us that
the distance from the Earth's center at that height is about

(1 / √0.66) = 1.23 times

the Earth's radius, so the height is about 910 miles above the surface.

7 0

3 years ago

(a) A submarine descends to a depth of 70 m below the surface of water. The density of the water is 1050 kg/m3. Atmospheric pres

timurjin [86]

Answer:

sttouyietETwe2e664yrwtwwteuwtrwruwuuwwuwtwuw7w7w5w7w772253536464647

5 0

2 years ago