Answer :
Using the relation between acceleration, velocity and distance, it is found that:
a) The differential equation for the velocity is: [tex]\frac{dV}{dt} = 0.5[/tex]
b) The formula for the velocity is given by: V(t) = 0.5t + 6.
c) The initial value problem to find the position is: [tex]\frac{dS}{dt} = 0.5t + 6, S(0) = 0[/tex]
d) The formula for the position of the car is S(t) = 0.25t² + 6t.
e) The car is at a velocity of 21 m/s after 30 seconds.
What is the relation between acceleration, velocity and distance?
We have that:
- The velocity is the integral of the acceleration.
- The position is the integral of the velocity.
For item a, we have that the acceleration is constant of 0.5 m/s², hence the differential equation for the velocity is:
[tex]\frac{dV}{dt} = 0.5[/tex]
For item b, we apply separation of variables, and find the velocity as follows:
[tex]\frac{dV}{dt} = 0.5[/tex]
[tex]\int dV = \int 0.5 dt[/tex]
V(t) = 0.5t + v(0).
v(0) is the initial velocity and is the constant of integration, hence:
The formula for the velocity is given by: V(t) = 0.5t + 6.
For item c, we have that the position is the integral of the velocity, hence:
The initial value problem to find the position is: [tex]\frac{dS}{dt} = 0.5t + 6, S(0) = 0[/tex]
For item d, applying separation of variables, we have that:
[tex]\frac{dS}{dt} = 0.5t + 6[/tex]
[tex]\int dS = \int (0.5t + 6) dt[/tex]
S(t) = 0.25t² + 6t + s(0).
Since s(0) = 0, we have that:
The formula for the position of the car is S(t) = 0.25t² + 6t.
The velocity after 30 seconds is of:
v(30) = 6 + 0.5(30) = 21.
The car is at a velocity of 21 m/s after 30 seconds.
More can be learned about the relation between acceleration, velocity and distance at https://brainly.com/question/25749514
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