Answer :
Let's solve the given problem step by step.
You have a displacement function [tex]\( s \)[/tex] given by:
[tex]\[ s = \frac{1}{2} t^2 - 5t + 16 \][/tex]
where [tex]\( s \)[/tex] is the displacement in feet, and [tex]\( t \)[/tex] is the time in seconds.
### (a) Find the velocity function
To find the velocity function, we need to differentiate the displacement function with respect to [tex]\( t \)[/tex]. The velocity [tex]\( v \)[/tex] is given by the first derivative of [tex]\( s \)[/tex] with respect to [tex]\( t \)[/tex].
[tex]\[ s(t) = \frac{1}{2} t^2 - 5t + 16 \][/tex]
Taking the derivative:
[tex]\[ v(t) = \frac{d}{dt} \left( \frac{1}{2} t^2 - 5t + 16 \right) \][/tex]
Differentiating each term separately:
[tex]\[ v(t) = \frac{d}{dt} \left( \frac{1}{2} t^2 \right) - \frac{d}{dt} (5t) + \frac{d}{dt} (16) \][/tex]
Using basic differentiation rules, we get:
[tex]\[ v(t) = \frac{1}{2} \cdot 2t - 5 \cdot 1 + 0 \][/tex]
[tex]\[ v(t) = t - 5 \][/tex]
So the velocity function is:
[tex]\[ v(t) = t - 5 \][/tex]
### (b) Find the velocity at a specific time
To determine the velocity at a specific time, say [tex]\( t = a \)[/tex], simply substitute [tex]\( a \)[/tex] into the velocity function.
[tex]\[ v(a) = a - 5 \][/tex]
### (c) Find when the particle returns to its initial position (where [tex]\( s = 0 \)[/tex])
To find the time when the particle returns to its original position (i.e., when [tex]\( s(t) = 0 \)[/tex]), we solve the equation:
[tex]\[ \frac{1}{2} t^2 - 5t + 16 = 0 \][/tex]
This is a quadratic equation in the form of:
[tex]\[ 0.5 t^2 - 5t + 16 = 0 \][/tex]
Using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 0.5 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = 16 \)[/tex].
### (d) Find the velocity when the particle hits the surface
To find the velocity when the particle hits the surface, substitute the values back into the velocity function. This is the [tex]\( v(t) \)[/tex] we found:
[tex]\[ v(t) = t - 5 \][/tex]
Substitute the value of [tex]\( t \)[/tex] that we found in part (c).
In summary:
(a) The velocity function is:
[tex]\[ v(t) = t - 5 \][/tex]
(b) The velocity at a specific time [tex]\( t = a \)[/tex] is:
[tex]\[ v(a) = a - 5 \][/tex]
(c) Solve [tex]\( 0.5 t^2 - 5t + 16 = 0 \)[/tex] to find when the particle hits the surface.
(d) Substitute the result from part (c) into the velocity function [tex]\( v(t) = t - 5 \)[/tex] to find the velocity when the particle hits the surface.
You have a displacement function [tex]\( s \)[/tex] given by:
[tex]\[ s = \frac{1}{2} t^2 - 5t + 16 \][/tex]
where [tex]\( s \)[/tex] is the displacement in feet, and [tex]\( t \)[/tex] is the time in seconds.
### (a) Find the velocity function
To find the velocity function, we need to differentiate the displacement function with respect to [tex]\( t \)[/tex]. The velocity [tex]\( v \)[/tex] is given by the first derivative of [tex]\( s \)[/tex] with respect to [tex]\( t \)[/tex].
[tex]\[ s(t) = \frac{1}{2} t^2 - 5t + 16 \][/tex]
Taking the derivative:
[tex]\[ v(t) = \frac{d}{dt} \left( \frac{1}{2} t^2 - 5t + 16 \right) \][/tex]
Differentiating each term separately:
[tex]\[ v(t) = \frac{d}{dt} \left( \frac{1}{2} t^2 \right) - \frac{d}{dt} (5t) + \frac{d}{dt} (16) \][/tex]
Using basic differentiation rules, we get:
[tex]\[ v(t) = \frac{1}{2} \cdot 2t - 5 \cdot 1 + 0 \][/tex]
[tex]\[ v(t) = t - 5 \][/tex]
So the velocity function is:
[tex]\[ v(t) = t - 5 \][/tex]
### (b) Find the velocity at a specific time
To determine the velocity at a specific time, say [tex]\( t = a \)[/tex], simply substitute [tex]\( a \)[/tex] into the velocity function.
[tex]\[ v(a) = a - 5 \][/tex]
### (c) Find when the particle returns to its initial position (where [tex]\( s = 0 \)[/tex])
To find the time when the particle returns to its original position (i.e., when [tex]\( s(t) = 0 \)[/tex]), we solve the equation:
[tex]\[ \frac{1}{2} t^2 - 5t + 16 = 0 \][/tex]
This is a quadratic equation in the form of:
[tex]\[ 0.5 t^2 - 5t + 16 = 0 \][/tex]
Using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 0.5 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = 16 \)[/tex].
### (d) Find the velocity when the particle hits the surface
To find the velocity when the particle hits the surface, substitute the values back into the velocity function. This is the [tex]\( v(t) \)[/tex] we found:
[tex]\[ v(t) = t - 5 \][/tex]
Substitute the value of [tex]\( t \)[/tex] that we found in part (c).
In summary:
(a) The velocity function is:
[tex]\[ v(t) = t - 5 \][/tex]
(b) The velocity at a specific time [tex]\( t = a \)[/tex] is:
[tex]\[ v(a) = a - 5 \][/tex]
(c) Solve [tex]\( 0.5 t^2 - 5t + 16 = 0 \)[/tex] to find when the particle hits the surface.
(d) Substitute the result from part (c) into the velocity function [tex]\( v(t) = t - 5 \)[/tex] to find the velocity when the particle hits the surface.
