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nikitadnepr [17]
4 years ago
10

An object moves with constant acceleration 4.05 m/s2 and over a time interval reaches a final velocity of 11.2 m/s.

Physics
1 answer:
aleksandrvk [35]4 years ago
5 0

a) Displacement: 11.6 m

b) Distance: 11.6 m

c) Displacement: 11.6 m

d) Distance: 19.4 m

Explanation:

a)

The motion of the object is a uniformly accelerated motion, so we can use the following suvat equation to find the displacement:

v^2-u^2=2as

where

s is the displacement

u is the initial velocity

a is the acceleration

v is the final velocity

For the object in this problem,

u = 5.60 m/s

v = 11.2 m/s

a=4.05 m/s^2

Solving for s, we find the displacement:

s=\frac{v^2-u^2}{2a}=\frac{11.2^2-5.60^2}{2(4.05)}=11.6 m

b)

Since the motion of the object is in a straight line and the direction of the velocity does not change, the distance is equal to the displacement, therefore

d = 11.6 m

c)

We can find the displacement of the object in this case by using again the equation

v^2-u^2=2as

where

s is the displacement

u is the initial velocity

a is the acceleration

v is the final velocity

This time, we have:

u = -5.60 m/s

v = 11.2 m/s

a=4.05 m/s^2

Solving for s, we find the displacement:

s=\frac{v^2-u^2}{2a}=\frac{11.2^2-(-5.60)^2}{2(4.05)}=11.6 m

d)

In this case, the velocity of the object changes direction: this means that for part of the trip the object moves in the negative direction, and for part of the trip it moves in the positive direction. Therefore, the distance is not equal to the displacement.

In order to find the distance, we have to calculate first the distance covered when the object is moving in the negative direction; so, from when it has velocity of

u = -5.60 m/s

to when it has a velocity of

v = 0

Therefore,

d_1=\frac{v^2-u^2}{2a}=\frac{0-(-5.60)^2}{2(4.05)}=3.9 m

Then, we have to add the distance covered from when the object has velocity

u = 0

to when it has velocity of

v = 11.2 m/s

Therefore,

d_2=\frac{v^2-u^2}{2a}=\frac{11.2^2-(0)^2}{2(4.05)}=15.5 m

Therefore, the total distance covered is

d=d_1+d_2=3.9 + 15.5=19.4 m

Learn more about distance and displacement:

brainly.com/question/3969582

#LearnwithBrainly

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5 0
2 years ago
A motorist is traveling at 20 m/s. He is 60 m from a stoplight when he sees it turn yellow. Is reaction time, before stepping on
Gnesinka [82]
V₀ = V₁ + 2ax
V₀ = final velocity which is 0 m/s
V₁ = initial velocity which is 20 m/s
x = distance which is 60-(0.5 x 20) =50m
     this is because his reaction time is 0.5 sec so he traveled 10m before stepping on the break paddle.

a= (V₀-V₁) / 2x
  = (0-20) / 2*50
  = -0.2m/s
the negative is because it is a deceleration speed hence it is 0.2m/s
5 0
4 years ago
I'm completely stumped as to how to do this.
worty [1.4K]

Explanation:

You have already determined the components of the known forces so I won't repeat your work here. Since the resultant force \vec{\textbf{R}} and F1 are completely along the x-axis, we can conclude that

F_{2y} = F_{3y} \Rightarrow F_{3y} = F_3\cos{\theta} = 250\:\text{lb}

We can now solve for the magnitude of F_3:

F_3 = \sqrt{F_{3x}^2 + F_{3y}^2} = \sqrt{(467)^2 + (250)^2}

\:\:\:\:=529.7\:\text{lb}

The angle \theta is then

\tan{\theta} = \dfrac{F_{3y}}{F_{3x}} = \dfrac{250}{467}

or

\theta = 49.2°

6 0
3 years ago
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