If a rock is thrown upward on the planet Mars with a velocity of 11 m/s, its height above the ground (in meters) after [tex]\( t \)[/tex] seconds is given by [tex]\( H = 11t - 1.86t^2 \)[/tex].

(a) Find the velocity (in m/s) of the rock after 1 second.

(b) Find the velocity (in m/s) of the rock when [tex]\( t = a \)[/tex].

(c) When (in seconds) will the rock hit the surface? (Round your answer to one decimal place.)

(d) With what velocity (in m/s) will the rock hit the surface?


The displacement (in feet) of a particle moving in a straight line is given by [tex]\( s = \frac{1}{2}t^2 - 5t + 16 \)[/tex], where [tex]\( t \)[/tex] is measured in seconds.



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.