Physics, 06.09.2019 17:30 yasmincx

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) list the knowns in this problem. (c) how far does the car travel in those 12.0 s? to solve this part, first identify the unknown, then indicate how you chose the appropriate equation to solve for it. after choosing the equation, show your steps in solving for the unknown, check your units, and discuss whether the answer is reasonable. (d) what is the car’s final velocity? solve for this unknown in the same manner as in (c), showing all steps explicitly

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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:

$V_{f}=28.8m/s$

While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) draw a

Answer from: Quest

f=ma newton

f=100x2.5=250n

Answer from: Quest

Initial velocity, u = 4.9 m/sfinal velocity = v m/sinitial height, h₁ = 245 m final height, h₂ = 0a = -9.8 m/s²time taken = t secondss = ut + 0.5at²⇒ (h₂-h₁) = ut + 0.5at²⇒ 0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get t = 7.59sv = u + at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is 69.48 m/s downward

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While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) draw a...

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