Answer:
The total distance at 7 s is:
[tex]x_{tot}=27\: m[/tex]
Explanation:
Distance due to the force
We can use second Newton's law to find the acceleration.
[tex]F=ma[/tex]
[tex]a=\frac{F}{m}=\frac{9}{5}=1.8\: m/s^{2}[/tex]
Now, using the kinematic equation we will find the distance during this interval of time. Let's recall that the initial velocity is zero.
[tex]x_{1}=0.5at_{1}^{2}[/tex]
[tex]x_{1}=0.5(1.8)(3)^{2}[/tex]
[tex]x_{1}=8.1\: m[/tex]
In the second part of the motion, the object moves at a constant velocity, as long as there is no friction between the object and the floor.
First, we need to find the final velocity of the first interval
[tex]v=v_{i}+at_{1}=0+(1.8)3=5.4\: m/s[/tex]
So the second distance will be:
[tex]x_{2}=vt_{2}=5.4*4=21.6\: m[/tex]
Therefore, the total distance is:
[tex]x_{tot}=x_{1}+x_{2}=5.4+21.6=27\: m[/tex]
I hope it helps you!
The total distance at 7 s is: [tex]x_{total}=27\ m[/tex]
Force is an external agent applied on any object to displace it from its position. Force is a vector quantity, so with magnitude it also requires direction. Direction is necessary to examine the effect of the force and to find the equilibrium of the force.
Distance due to the force
We can use second Newton's law to find the acceleration.
[tex]F=ma[/tex]
[tex]a=\dfrac{F}{m}=\dfrac{9}{5}=1.8\ \frac{m}{s^2}[/tex]
Now, using the kinematic equation we will find the distance during this interval of time. Let's recall that the initial velocity is zero.
[tex]x_1=0.5at_1^2[/tex]
[tex]x_1=0.5(1.8)(3^2)[/tex]
[tex]x_1=8.1\ m[/tex]
In the second part of the motion, the object moves at a constant velocity, as long as there is no friction between the object and the floor.
First, we need to find the final velocity of the first interval
[tex]v=v_i+at_1=0+(1.8)(3)=5.4 \frac{m}{s}[/tex]
So the second distance will be:
[tex]x_2=vt_2=5.4\times 4=21.6\ m[/tex]
Therefore, the total distance is: [tex]x_{total}=27\ m[/tex]
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