There are (one can say) three coequal theories of motion for a single particle: Newton's second law, stating that the total force on an object causes its acceleration; the work–kinetic energy theorem, stating that the total work on an object causes its change in kinetic energy; and the impulse–momentum theorem, stating that the total impulse on an object causes its change in momentum. In this problem, you compare predictions of the three theories in one particular case. A 2.40-kg object has velocity 7.00ĵ m/s. Then, a constant net force 14.0î N acts on the object for 3.50 s.a) Calculate the object's final velocity, using the impulse–momentum theorem.
vf= m/s
b) ) Calculate its acceleration from a=(vf-vi)/Δt
a= m/s^2
c) Calculate its acceleration from a=ΣF/m
a= m/s^2
d) Find the object's vector displacement from Δr=vit+(1/2)at^2
Δr= m

Therefore, here we apply the impulse-momentum theorem only on the horizontal motion of the particle, since the force acts only in that direction. The impulse must be equal to the change in momentum along the x-direction, so we can write:

[tex]F \Delta t = m(v_x - u_x)[/tex]

where

F = 14.0 N is the force

[tex]\Delta t = 3.50 s[/tex] is the time during which the force is applied

m = 2.40 kg is the mass of the object

[tex]v_x[/tex] is the final x-velocity of the object

[tex]u_x = 0[/tex] is the initial x-velocity of the object

Solving for [tex]v_x[/tex],

[tex]v_x = \frac{F \Delta t}{m}=\frac{(14.0)(3.50)}{2.40}=20.4 m/s[/tex]

And so, the final velocity of the object will be

[tex]v=(7.00 i+20.4j) m/s[/tex]

b)

The acceleration of the object is given by

[tex]a=\frac{v_f-v_i}{\Delta t}[/tex]

where

[tex]v_f[/tex] is the final velocity

[tex]v_i[/tex] is the initial velocity

[tex]\Delta t[/tex] is the time interval

Since acceleration is a vector, we have to calculate its components. In the x-direction we have:

[tex]v_f = 20.4 m/s\\v_i = 0[/tex]

So, the x-component of the acceleration is

[tex]a_x=\frac{20.4-0}{3.50}=5.8 m/s^2[/tex]

In the y-direction, we have

[tex]v_f = 7.00 m/s\\v_i = 7.00 m/s[/tex]

So, the y-component of the acceleration is

[tex]a_y=\frac{7.00-7.00}{3.50}=0[/tex]

So, the acceleration is

[tex]a=(5.8 i) m/s^2[/tex]

c)

Here we want to calculate the acceleration using

[tex]a=\frac{F}{m}[/tex]

In this case we have:

F = 14.0i N is the force (along the x-direction)

m = 2.40 kg is the mass

Since the force is only along the x-direction, the acceleration has only component on this direction, and therefore it is given by

[tex]a_x = \frac{14.0}{2.40}=5.8 m/s^2[/tex]

So, the acceleration is

[tex]a=(5.8 i) m/s^2[/tex]

d)

The displacement of the object is given by

[tex]\Delta r = v_i t + \frac{1}{2}at^2[/tex]

Since it is a vector, we have to find the components of the displacement along the two directions.

Along the x-direction, we have:

[tex]v_i = 0\\a = 5.8 m/s^2[/tex]

So, the x-component of the displacement is

[tex]\Delta r_x = 0 + \frac{1}{2}(5.8)(3.50)^2=35.5 m[/tex]

In the y-direction, we have

[tex]v_i = 7.00\\a = 0 m/s^2[/tex]

So, the y-component of the displacement is

[tex]\Delta r_y = (7.00)(3.50) + 0=24.5 m[/tex]

Therefore, the object's vector displacement is

[tex]\Delta r = (35.5 i +24.5j) m[/tex]

Learn more about acceleration:

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brainly.com/question/2562700

And momentum-impulse:

brainly.com/question/9484203

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