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3 years ago

Calculate the acceleration of Josh riding his bicycle in a straight line that speeds up from 4 m/s to 6 m/s in 5 seconds

Physics

2 answers:

Anton [14]3 years ago

8 0

Answer:

$a=0.4m/s^2$

Explanation:

The formula for calculating acceleration is as follows:

$a=\frac{v_{f}-v_{i}}{t}$

where $v_{f}$ is the final velocity, $v_{i}$ the initial velocity, an $t$ is the time in wich the velocity changed.

For this case:

$v_{f}=6m/s$

$v_{i}=4m/s$

$t=5s$

Thus the acceleration is:

$a=\frac{6m/s-4m/s}{5s}$

$a=\frac{2m/s}{5s}$

$a=0.4m/s^2$

The acceleration of Josh is: $a=0.4m/s^2$

Varvara68 [4.7K]3 years ago

3 0

Answer:

U=4m/s

V=6m/s

T=5s

Acceleration =v-u/t

6-4/5

=2/5 metre/second(square).

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In science, a broad idea that has been repeatedly verified so as to give scientists great confidence that it represents reality is called "a theory".

Explanation:

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2 years ago

A stone is dropped from the

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Height=h=500m
Acceleration=g=10m/s^2
Initial velocity=u=0
Speed of sound=c=340m/s
TIME TAKEN BY STONE TO HIT WATER=t
Time taken by sound to hear back=T

Now

$\\ \sf\longmapsto h=ut+\dfrac{1}{2}gt^2$

$\\ \sf\longmapsto h=0t+\dfrac{1}{2}10t^2$

$\\ \sf\longmapsto 500=5t^2$

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Now

$\\ \sf\longmapsto h=cT$

$\\ \sf\longmapsto T=\dfrac{h}{c}$

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2 years ago

While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)

ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration= $2.40m/s^{2}$ , c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = $2.40m/s^{2}$

time = 12.0 s

initial position = 0

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

$x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}$

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

$x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}$

after cancelling the pertinent units, I get that my answer will be given in meters. So I get:

$x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}$

which solves to:

$x=172.8m$

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

$a=\frac{V_{f}-V_{0} }{t}$

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

$at=V_{f}-V_{0}$

and finally I can add $V_{0}$ to both sides so I get:

$V_{f}=at+V_{0}$

and now I can proceed and substitute the known values:

$V_{f}=at+V_{0}$

$V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0$

which solves to:

$V_{f}=28.8m/s$

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