Answer:
No Time
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Mass of the car = 1200 kg
Mass of the truck = 2100 kg
Total mass of car and truck = 2100 + 1200 = 3300 kg
Since, the car pushes the truck. Hence, they will move together and will have same acceleration.
Let the acceleration be a.
According to Newton's second law:
F(net) = ma
F = 4500 N
4500 = 3300 × a
![a = \frac{4500}{3300}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B4500%7D%7B3300%7D)
a = 1.36 m/s^2
Let the force applied by the car on truck be F.
F = F(net) on the truck
F = ma
F = 2100 × 1.36
F = 2856 N
Hence, the force applied by the car on the truck is 2856 N
Answer:
The constant value is ![k\approx 0.17\,m/s^4](https://tex.z-dn.net/?f=k%5Capprox%200.17%5C%2Cm%2Fs%5E4)
The net displacement is ![D=53.64\, m](https://tex.z-dn.net/?f=D%3D53.64%5C%2C%20m)
Explanation:
If the acceleration as a function of time is given
then, first of all, knowing that the units of acceleration should be
we should have
where
stands for The dimension of k (these are just the units of k in a less formal way of saying it.
On the other hand we have only information about the velocity, but we only have the acceleration function, it turns out we can integrate the expression of acceleration in order to obtain the velocity as a function of time:
where
as a constant of integration which should have units of
in order to be consistent with the fact that it is a velocity function, it is therefore natural to think of
as the initial velocity of the the particle.
Let's now get our hands dirty by integrating ![a(t)](https://tex.z-dn.net/?f=a%28t%29)
.
By having the velocity as a function of time we can now use the conditions given at t=0 and t=6.
At t=0 we have:
![v(0)=-k\frac{0^3}{3}+v_0=12\implies v_0=12\, m/s](https://tex.z-dn.net/?f=v%280%29%3D-k%5Cfrac%7B0%5E3%7D%7B3%7D%2Bv_0%3D12%5Cimplies%20v_0%3D12%5C%2C%20m%2Fs)
At t=6 the particle start reversing direction, that means at that very instant it velocity should be zero in order to start traveling the other way. This can only mean the following
.
We have a full description now of the acceleration and the velocity function. In order to get the net displacement we need to integrate the velocity function
![x(t)=\int v(t)=-\frac{k}{3}\int 3 dt+\int v_0 dt=-\frac{k}{12}t^4+v_0\cdot t+x_0](https://tex.z-dn.net/?f=x%28t%29%3D%5Cint%20v%28t%29%3D-%5Cfrac%7Bk%7D%7B3%7D%5Cint%203%20dt%2B%5Cint%20v_0%20dt%3D-%5Cfrac%7Bk%7D%7B12%7Dt%5E4%2Bv_0%5Ccdot%20t%2Bx_0%20)
Where
is the initial displacement. If we subtract
on both sides we get the net displacement or distance traveled
![x(t)-x_0=D=-\frac{kt^4}{12}+12t](https://tex.z-dn.net/?f=x%28t%29-x_0%3DD%3D-%5Cfrac%7Bkt%5E4%7D%7B12%7D%2B12t)
Plugging the value of 6 above gives us the net displacement
.
A=v-v0 divided by t equals delta(the triangle) V divided by delta t
I would say a because it just sounds right. I would've helped u if u had a real question question.