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Mashutka [201]
3 years ago
9

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 49 m in front of you. You reaction time

before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 10m/s2.
a. How much distance is between you and the deer when you come to a stop?
b. What is the maximum speed you could have and still not hit the deer?
Physics
1 answer:
Leno4ka [110]3 years ago
4 0

Answer:

v = 26.7 m/s

Explanation:

Given,

speed of the car = 20 m/s

distance between the car and the deer = 49 m

time taken to press the brake = 0.50 s

maximum deceleration of the car = 10 m/s²

Now,

distance travel by the car in 0.5 s = u x t = 20 x 0.5 = 10 m

distance travel by the car after the break is pressed

Using equation of motion

v² = u² + 2 a s

0² = 20² - 2 x 10 x s

s = 20 m

Total distance travel by the car = 20 + 10 = 30 m

Distance between deer and car = 49-30 = 19 m.

b. Maximum speed a car could have

Distance travel by the car in reaction time = v' x 0.5

v' is the maximum speed of the car.

maximum distance car can cover = 49 - 0.5 v'

Now, Using equation of motion

v² = u² + 2 a s

0² =v'² - 2 x 10 x (49- 0.5 x v')

v'² +10 v' -980 = 0

By solving

v = 26.7 m/s

Hence, maximum speed of the car can be 26.7 m/s

 

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60 km

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For an object (or a person, such as in this case) moving at constant speed, the speed is equal to the ratio between the distance travelled and the time taken:

where

v is the speed

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t is the time taken

In this case, we have:

v = 120 km/h is the speed

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Go outside on a clear night, wait 15 minutes for your eyes to adjust to the dark, and look carefully at the brightest stars. Som
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blue star is hotter than red

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What will be the restoring force if a spring with a spring constant of 45 newtons per meter is pulled 0.03 meters in the downwar
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To calculate for the force in a spring, we use Hooke's Law which relates force and the displacement of the spring. It is said that the force is directly proportional to the displacement. So, it will have the equation F = kx where k is a constant and it is the spring constant.

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8 0
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5.
igor_vitrenko [27]

Answer:

a=40\ m/s^2

Explanation:

Given that,

Initial speed of a shuttlecock, u = 30 m/s

Final speed of the shuttlecock, v = 10 m/s

Time, t = 0.5 s

We need to find its average acceleration. The acceleration of an object is equal to the change in speed divided by time taken. It is given by :

a=\dfrac{v-u}{t}\\\\a=\dfrac{10-30}{0.5}\\\\a=-40\ m/s^2

So, the average acceleration of badminton shuttlecock is 40\ m/s^2.

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A boat with a horizontal tow rope pulls a waterskier. She skis off to the side, the rope makes an angle of 15° with the forward
dsp73

Answer:

<h3>13,976.23Joules</h3>

Explanation:

Workdone by the rope is expressed using the formula;

W = Fd sin(theta)

F is the tension in the rope = 180

d is the displacement = 300m

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Substitute the given parameters into the formula;

W = 180(300)sin15

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W = 13,976.23

Hence the workdone by the rope is 13,976.23Joules

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