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kifflom [539]
3 years ago
15

A car traveling at 20 m/s when the driver sees a child standing in the road. He takes 0.80 s to react, then steps on the brakes

and slows at 7.0 m/s2. How far does the car go before it stops?
Physics
1 answer:
mr Goodwill [35]3 years ago
3 0

When driver see the child standing on road his speed is 20 m/s

So here at that instant his reaction time is 0.80 s

He will cover a total distance given by product of speed and time

d_1 = v* t

d_1 = 20 * 0.8

d_1 = 16 m

now after this he will apply brakes with acceleration a = 7 m/s^2

so the distance covered before it stop is given by

v_f^2 - v_i^2 = 2 a d

0 - 20^2 = 2*(-7)*d_2

d_2 = 28.6 m

so the total distance covered by it

d = d_1 + d_2

d = 16 + 28.6 = 44.6 m

<em>so it will cover a total distance of 44.6 m</em>

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The site from which an airplane takes off is the origin. The X axis points east, the y axis points straight up. The position and
Contact [7]

Answer:

d = 3.5*10^4 m

Explanation:

In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:

\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m

The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:

d=\sqrt{(x-x_o)^2+(y-y_o)^2}   (1)

where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):

d=\sqrt{(1.12*10^3-0)^2+(3.45*10^4-0)^2}=3.45*10^4m

hence, the displacement of the airplane is 3.45*10^4 m

6 0
3 years ago
1. Two forces act on a box as follows: F1 = 100 N at 01 = 170° and F2 = 75 N
lilavasa [31]

Answer:

a)  F = 64.30 N,  b) θ = 121.4º

Explanation:

Forces are vector quantities so one of the best methods to add them is to decompose each force and add the components

let's use trigonometry

Force F1

          sin 170 = F_{1y} / F₁

          cos 170 = F₁ₓ / F₁

          F_{1y} = F₁ sin 170

          F₁ₓ = F₁ cos 170

          F_{1y} = 100 sin 170 = 17.36 N

          F₁ₓ = 100 cos 170 = -98.48 N

Force F2

          sin 30 = F_{2y} / F₂

          cos 30 = F₂ₓ / F₂

          F_{2y} = F₂ sin 30

          F₂ₓ = F₂ cos 30

          F_{2y} = 75 sin 30 = 37.5 N

          F₂ₓ = 75 cos 30 = 64.95 N

the resultant force is

X axis

          Fₓ = F₁ₓ + F₂ₓ

          Fₓ = -98.48 +64.95

          Fₓ = -33.53 N

Y axis

         F_y = F_{1y} + F_{2y}

         F_y = 17.36 + 37.5

         F_y = 54.86 N

a) the magnitude of the resultant vector

let's use Pythagoras' theorem

         F = Ra Fx ^ 2 + Fy²

         F = Ra 33.53² + 54.86²

         F = 64.30 N

b) the direction of the resultant

let's use trigonometry

        tan θ’= F_y / Fₓ

        θ'= tan^{-1}  \frac{F_y}{F_x}

        θ'= tan⁻¹ (54.86 / (33.53)

        θ’= 58.6º

this angle is in the second quadrant

The angle measured from the positive side of the x-axis is

        θ = 180 -θ'

        θ = 180- 58.6

        θ = 121.4º

5 0
3 years ago
The acceleration due to the earth's gravity, in si units, is 9.8 m/s2. in the absence of air friction, a ball is dropped from re
gogolik [260]
We don't know anything about the amount of distance it travels, but that's okay. The only equation we need here is 

velocity(final) = velocity(initial) + acceleration * time
vf = vi + (a * t)

The ball is dropped from rest, so vi = 0 m/s.
We want it so that the ball hits the ground with a final velocity of 60 m/s, so vf = 60 m/s. 
We are given the acceleration due to gravity, a = 9.8 m/s^2.
We are solving for the time, t = ?.

Now we just plug in the values.
vf = vi + (a * t)
60 m/s = 0 m/s + (9.8 m/s^2)*(t)

60 = 9.8t

60 / 9.8 = t

t = 6.122 s

Hopefully this is the right answer.

7 0
3 years ago
You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped a
AfilCa [17]

Answer:

Explanation:

a Downward acceleration of car A along the slope

= g sinθ - μ g cosθ

= g ( sinθ - μ  cosθ)

= 9.8 ( sin 12 - .6 x cos 12 )

= 9.8 x ( .2079 - .5869 )

= - 3.714 m / s²

So there will be deceleration

v² = u² - 2 a s

= 18² - 2 x 3.714 x 24

= 324 - 178

= 146

v = 12 .08 m /s

b )

In the second case , kinetic friction changes

downward acceleration

= g ( sinθ - μ  cosθ)

= 9.8 ( sin12 - .1 x cos 12 )

9.8 ( .2079  -  .0978 )

= 1.079 m /s

there will be reduced acceleration

v² = u² - 2 a s

= 18² +2 x1.079 x 24

= 324 + 52

= 376

v = 19.4  m /s

7 0
3 years ago
How does the egg drop project apply to newtons 3 laws of motion
iren2701 [21]
The reason why it relates to the newtons 3 laws of motion because the first law of motion states that every object will stay at rest unless it's moved by an unbalanced force which is your hand. The second one states that <span> the velocity of an object changes when it is subjected to an external force meaning it's used by the equation that is commonly used for which is F=M*A. The way it relates to the second law because you are adding force some way or another, the mass is the egg and the acceleration is the drop of the egg while it free falls. And the last one, for a reaction there is always an equal or opposite reaction and the opposite reaction is the floor because it's going against the egg causing it to crack. If it was with the egg, it would have a soft, smooth landing.</span>
8 0
4 years ago
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