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3 years ago

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A cheetah can run at a maximum speed 103 km/h and a gazelle can run at a maximum speed of 76.5 km/h. If both animals are running

at full speed, with the gazelle 99.3 m ahead, how long before the cheetah catches its prey?
Answer in units of s.

Part 2: The cheetah can maintain its maximum speed for only 7.5 s. What is the minimum distance the gazelle must be ahead of the cheetah to have a chance of escape? (After 7.5 s the speed of cheetah is less than that of the gazelle.)

Answer in units of m.

1 answer:

Katena32 [7]3 years ago

5 0

Answer:

1

$t_a = 13.49 \ s$

2

The distance is $D = 55.2 \ m$

Explanation:

From the question we are told that

The maximum speed of the cheetah is $v = 103 \ km/h = 28.61 \ m/s$

The maximum of gazelle is $u = 76.5 \ km/h = 21.25 \ m/s$

The distance ahead is $d = 99.3 \ m$

Let $t_a$ denote the time which the cheetah catches the gazelle

Gnerally the equation representing the distance the cheetah needs to move in order to catch the gazelle is

$v* t_a = d + u* t_a$

=> $28.61 t_a = 99.3 + 21.25t_a$

=> $7.36 t_a = 99.3$

=> $t_a = 13.49 \ s$

Now at t = 7.5 s

$7.5 v = D+ 7.5u$

=> $28.61 * 7.5 = D + 21.25* 7.5$

=> $7.36 * 7.5 =D$

=> $D = 55.2 \ m$

Hence the for the gazelle to escape the cheetah it must be 55.2 m

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A woman can row a boat at 5.60 km/h in still water. (a) If she is crossing a river where the current is 2.80 km/h, in what direc

katrin2010 [14]

Answer:

a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.

Explanation:

So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

$cos \theta = \frac {V_{river}}{V_{boat}}$

When solving for theta, we get that:

$\theta =cos^{-1} ( \frac {V_{river}}{V_{boat}})$

so now we can substitute the corresponding values:

$\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})$

Which yields:

$\theta = 60^{o}$

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

$V_{y}=5.60km/hr*cos(210^{o})$

$V_{y}=-4.85km/hr$

The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:

$t=\frac{y}{V_{y}}$

$t=\frac{5.60km}{4.85km/hr}=1.155hr$

part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

$V_{ds}=V_{river}+V{boat}$

$V_{ds}=2.80km/hr+5.60km/hr=8.40km/hr$

and now up stream:

$V_{us}=V_{boat}-V{river}$

$V_{us}=5.60km/hr-2.80km/hr=2.80km/hr$

Once we got these two velocities we will now need to find the time to take each trip:

time down stream:

$t_{ds}=\frac{x}{v_{ds}}$

$t_{ds}=\frac{2.80km}{8.40km/hr}=0.333hr$

and the time up stream:

$t_{us}=\frac{x}{v_{us}}$

$t_{us}=\frac{2.80km}{2,80km/hr}=1hr$

so the total time will be:

$t_{ds}+t_{us}=0.333hr+1hr=1.333hr$

d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:

t=1.333hr

e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)

f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

$t=\frac{d}{v_{boat}}$

$t=\frac{5.60km}{5.60km/hr}$

so

t= 1hr

4 0

3 years ago

Read 2 more answers

A 0.45kg baseball is pitched towards home plate at 20 m/s. The ball is hit back towards the pitcher with a speed of 30 m/s. What

Inessa05 [86]

Answer:

4.5kgm/s

Explanation:

Change in momentum is expressed as

Change in momentum = m(v-u)

M is the mass

V is the final velocity

u is the initial velocity

Given

m=0.45kg

v = 30m/s

u = 20m/s

Substitute

Change in momentum = 0.45(30-20)

Change in momentum = 0.45×10

Change in momentum = 4.5kgm/s

3 0

3 years ago

brainly a child on a swing set swings back and forth with a period of 3.3 s and an amplitude of 25°. what is the maximum speed o

beks73 [17]

Maximum speed of the child as she swings is 2.23 m/s.

<h3>Step by Step Calculation:</h3>

T=3.3 s is the oscillation's time period.

The swing's greatest angle is 25° (max).

The swing's bottom will have the following kinetic energy:

k=12mv2...........(1)

The mass in this situation is m, and the speed is v.

The potential energy change is expressed as,

∆U=mgL1-cosθmax...............(2)

Here, L is a string's length and g is the acceleration caused by gravity. L is given as,

L=gT24π2

Combine equation (1) with (2)

12mv2=mgL1-cos, maxv=2g, maxv=gT24, maxv=g2T22, maxv=9.8 m/s

22.33 m/s, 22.31 s22.31 cos25°

Therefore, the child's top speed is 2.23 m/s.

<h3>What is Oscillation ?</h3>

The process of any quantity or measure fluctuating repeatedly about its equilibrium value in time is known as oscillation.
A periodic change in a substance's value between two values or around its central value is another way to define oscillation.

To learn more about Oscillation refer to:

brainly.com/question/28312746

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6 0

1 year ago

what is the force constant, in newtons per meter, needed to produce a period of 0.45 s for a 0.011-kg mass on the spring?

topjm [15]

The force constant is 2.145 N/m.

<h3>What is spring constant?</h3>

The spring constant is the force required to stretch or compress a spring divided by the distance traveled by the spring. It is used to determine whether a spring is stable or unstable.

K is the proportionality constant, also known as the 'spring constant.' Hooke's law (F = -kx) specifies stiffness and strength via the k variable. The greater the value of k, the greater the force required to stretch an object to a given length.

Using the relation;

T = 2π√m/k

T = time period = 0.45 s

m = mass of object in kilograms = 0.011kg

k = spring constant

To find k based on the formula,

k = 4 × (3.142)^2 × 0.011 / (0.45 )^2

k = 2.145 N/m

Therefore the force constant is 2.145 N/m.

To learn more about force refer to :

brainly.com/question/12785175

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8 0

1 year ago

A hot air balloon is moving vertically upwards at a velocity of 3m/s. A sandbag is dropped when the balloon reaches 150m. How lo

gregori [183]

This is a perfect opportunity to stuff all that data into the general equation for the height of an object that has some initial height, and some initial velocity, when it is dropped into free fall.

   H(t) = (H₀) + (v₀ T) + (1/2 a T²)

Height at any time 'T' after the drop =

(initial height) +
      (initial velocity) x (T) +
   (1/2) x (acceleration) x (T²) .

For the balloon problem ...

-- We have both directions involved here, so we have to define them:

   Upward = the positive direction

   Initial height = +150 m
   Initial velocity = + 3 m/s

   Downward = the negative direction

   Acceleration (of gravity) = -9.8 m/s²

Height when the bag hits the ground = 0 .

   H(t) = (H₀) + (v₀ T) + (1/2 a T²)

   0 = (150m) + (3m/s T) + (1/2 x -9.8 m/s² x T²)

   -4.9 T² + 3T + 150 = 0

Use the quadratic equation:

   T = (-1/9.8) [ -3 plus or minus √(9 + 2940) ]

   = (-1/9.8) [ -3 plus or minus 54.305 ]

   = (-1/9.8) [ 51.305 or -57.305 ]

      T = -5.235 seconds or 5.847 seconds .

(The first solution means that the path of the sandbag is part of
the same path that it would have had if it were launched from the
ground 5.235 seconds before it was actually dropped from balloon
while ascending.)

Concerning the maximum height ... I don't know right now any other
easy way to do that part without differentiating the big equation.
So I hope you've been introduced to a little bit of calculus.

H(t) = (H₀) + (v₀ T) + (1/2 a T²)

   H'(t) = v₀ + a T

The extremes of 'H' (height) correspond to points where h'(t) = 0 .

Set    v₀ + a T = 0

+3 - 9.8 T = 0

Add 9.8 to each side: 3   = 9.8 T

Divide each side by 9.8 :   T = 0.306 second

That's the time after the drop when the bag reaches its max altitude.

Oh gosh ! I could have found that without differentiating.

- The bag is released while moving UP at 3 m/s .

- Gravity adds 9.8 m/s of downward speed to that every second.
So the bag reaches the top of its arc, runs out of gas, and starts
falling, after
   (3 / 9.8) = 0.306 second .

At the beginning of that time, it's moving up at 3 m/s.
At the end of that time, it's moving with zero vertical speed).
Average speed during that 0.306 second = (1/2) (3 + 0) = 1.5 m/s .

Distance climbed during that time = (average speed) x (time)

   = (1.5 m/s) x (0.306 sec)

   = 0.459 meter (hardly any at all)

   But it was already up there at 150 m when it was released.

It climbs an additional 0.459 meter, topping out at 150.459 m,
then turns and begins to plummet earthward, where it plummets
to its ultimate final 'plop' precisely 5.847 seconds after its release.

We can only hope and pray that there's nobody standing at
Ground Zero at the instant of the plop.

I would indeed be remiss if were to neglect, in conclusion,
to express my profound gratitude for the bounty of 5 points
that I shall reap from this work. The moldy crust and tepid
cloudy water have been delicious, and will not soon be forgotten.

6 0

3 years ago