Answer: [tex]m \frac{d}{dt}v_{(t)}[/tex]
Explanation:
In the image attached with this answer are shown the given options from which only one is correct.
The correct expression is:
[tex]m \frac{d}{dt}v_{(t)}[/tex]
Because, if we derive velocity [tex]v_{t}[/tex] with respect to time [tex]t[/tex] we will have acceleration [tex]a[/tex], hence:
[tex]m \frac{d}{dt}v_{(t)}=m.a[/tex]
Where [tex]m[/tex] is the mass with units of kilograms ([tex]kg[/tex]) and [tex]a[/tex] with units of meter per square seconds [tex]\frac{m}{s}^{2}[/tex], having as a result [tex]kg\frac{m}{s}^{2}[/tex]
The other expressions are incorrect, let’s prove it:
[tex]\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}[/tex] This result has units of [tex]kg\frac{m}{s}[/tex]
[tex]m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m[/tex] This result has units of [tex]kg[/tex]
[tex]m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C[/tex] This result has units of [tex]kgm^{2}[/tex] and [tex]C[/tex] is a constant
[tex]m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m[/tex] This result has units of [tex]kg[/tex]
[tex]m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m[/tex] This result has units of [tex]kg[/tex]
[tex]\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C[/tex] This result has units of [tex]kg \frac{m^{3}}{s^{3}}[/tex] and [tex]C[/tex] is a constant
[tex]m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C[/tex] This result has units of [tex]kg \frac{m^{2}}{s^{4}}[/tex] and [tex]C[/tex] is a constant
[tex]\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0[/tex] because [tex]v_{(x)}[/tex] is a constant in this derivation respect to [tex]t[/tex]
[tex]m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C[/tex] This result has units of [tex]kg \frac{m^{2}}{s^{2}}[/tex] and [tex]C[/tex] is a constant
