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leonid [27]
3 years ago
7

A wildebeest calf is cruising at its top speed of v= 10 m/s when it passes over a sleeping cheetah. By the time the cheetah stan

ds up and begins pursuit, the wildebeest is d= 7.0 ma head of the cheetah. If the cheetah is able to maintain a constant acceleration of a= 9.5 m/s^2until it catches the wild-beast, then how much time t must pass before the cheetah catches up to the wildebeast? (Hint: you will need to usethe quadratic formula.) For the limits check, investigate what happens to t as the initial separation distance approaches zero.
Physics
1 answer:
Gre4nikov [31]3 years ago
6 0

The time when the cheetah catches up with the widebeest is 1.53 s

If the initial separation distance approaches zero, it will take the cheetah 1.05 s to catch the widebeest.

The given parameters;

  • speed of the wildebeest calf, Vw = 10 m/s
  • distance traveled by the calf before the cheetah stands up = 7 m
  • constant acceleration of the cheetah, a = 9.5 m/s²

Let the speed of the cheetah = Vc

let the time the cheetah catches up with the wildebeest = t

a = \frac{V_c}{t}

V_c = at

Apply relative velocity formula to determine the time when the cheetah catches up with the widebeest;

Assuming the wildebeest and the cheetah are running in the same direction;

(V_c - V_w) t = 7 \\\\(at - V_w) t = 7\\\\(9.5t - 10)t = 7\\\\9.5t^2 - 10t = 7\\\\9.5t^2 -10t-7 = 0\\\\solve \ the \ quadratic \ equation \ using \ formula \ method;\\\\a = 9.5 \ b= -10 \ , c =-7\\\\t = \frac{-b \ \ +/- \ \ \sqrt{b^2-4ac} }{2a} \\\\t = \frac{-(-10)\ \ +/- \ \ \sqrt{(-10)^2-4(9.5\times -7)} }{2(9.5)}\\\\t = \frac{10 \ \ +/- \ \ 19.13}{19} \\\\t = 1.53 \ s

The time when the cheetah catches up with the widebeest is 1.53 s

If the initial separation distance approaches zero;

(V_c - V_w) t = 0\\\\(at - V_w) t = 0\\\\(9.5t - 10)t = 0\\\\9.5t^2 - 10t = 0\\\\9.5t^2 = 10t\\\\9.5t = 10\\\\t = \frac{10}{9.5} = 1.05 \ s

Thus, if the initial separation distance approaches zero, it will take the cheetah 1.05 s to catch the widebeest.

Lear more here: brainly.com/question/24430414

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Doc Brown holds on to the end of the minute hand of the clock atop city hall. The tangential velocity of the minute hand is 0.41
alexdok [17]

The Professor's centripetal acceleration is 0.044 m/s²

Centripetal acceleration is the acceleration of an object moving in circular motion. It is usually directed towards the center of the rotation.

It is given by:

a = v²/r

where v is the velocity and r is the radius.

Given that the radius (r) = 4 m, velocity (v) = 0.419 m/s, hence:

a = v²/r = 0.419²/4 =  0.044 m/s²

The Professor's centripetal acceleration is 0.044 m/s²

Find out more at: brainly.com/question/6082363

3 0
3 years ago
24) a plane takes off at 7:00 pm and lands at 1:00 am after travelling a distance of 4818km. what is the minimum number of times
12345 [234]

Answer:

Twice

Step-by-Step Explanation:

Time between 7:00 PM and 1:00 AM: 6 hours

Distance: 4818km

Since the distance is 4818km, and the time is 6 hours, you divide 4818 by 6.

803.0000015999 km/h.

The average speed is 803 km/h

Which considering the ideal case scenario if the plane starts at 0 reaches the speed of 803 and the end reduces its speed from 803 to 0. This means we have come across the value of 800 at least twice. Hence, the plane was travelling at a speed of 800 km/h at least 2 times.

7 0
1 year ago
A ball is thrown straight up. What are the velocity and acceleration of the ball at the highest point in its path?
zubka84 [21]

Answer:

b. v = 0, a = 9.8 m/s² down.

Explanation:

Hi there!

The acceleration of gravity is always directed to the ground (down) and, near the surface of the earth, has a constant value of 9.8 m/s². Since the answer "b" is the only option with an acceleration of 9.8 m/s² directed downwards, that would solve the exercise. But why is the velocity zero at the highest point?

Let´s take a look at the height function:

h(t) = h0 + v0 · t + 1/2 g · t²

Where

h0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity

Notice that the function is a negative parabola if we consider downward as negative (in that case "g" would be negative). Then, the function has a maximum (the highest point) at the vertex of the parabola. At the maximum point, the slope of the tangent line to the function is zero, because the tangent line is horizontal at a maximum point. The slope of the tangent line to the function is the rate of change of height with respect to time, i.e, the velocity. Then, the velocity is zero at the maximum height.

Another way to see it (without calculus):

When the ball is going up, the velocity vector points up and the velocity is positive. After reaching the maximum height, the velocity vector points down and is negative (the ball starts to fall). At the maximum height, the velocity vector changed its direction from positive to negative, then at that point, the velocity vector has to be zero.

8 0
3 years ago
The graph represents the force applied on an 3.00kg crate while it moved 5.0m. A. How much total work is done on the crate? B. I
jeka57 [31]

a. We can calculate the amount of work by calculating the area under the graph.

first area (rectangular): 2.5 x 6 = 15

second area(trapezoid): 1/2 x (6+10) x 2.5 =20

total work done: 35 J

b. the force was first applied = 6 N

F = m.a

a = 6 : 3 = 2 m/s²

vf²=vi²+2as

vf²=6²+2.2.5

vf²=56

vf=7.5 m/s

8 0
3 years ago
When sedimentary rock is exposed to heat and pressure, what does it change into?
patriot [66]
Metamorphic rock this possess often occurs in the mantle
7 0
3 years ago
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