Question

A car's velocity is modeled by V(t)=0.5t^2 -8t +24 ,where the velocity is in feet per second and time is in seconds. When does the car come to a complete stop
12 seconds
8 seconds
4 seconds
2 seconds

Answer 1

The velocity of the car is the rate of change of distance over time.

The car comes to a complete stop at 12 seconds

The velocity function is given as:

$\mathbf{V(t) = 0.5t^2 - 8t + 24}$

When the car comes to a complete stop, V(t) = 0.

So, we have:

$\mathbf{0.5t^2 - 8t + 24 = 0}$

Expand

$\mathbf{0.5t^2 - 6t - 2t + 24 = 0}$

Factorize

$\mathbf{0.5t(t - 12) - 2(t - 12) = 0}$

Factor out t - 12

$\mathbf{(0.5t - 2)(t - 12) = 0}$

Split

$\mathbf{(0.5t - 2) =0\ or\ (t - 12) = 0}$

Solve for t

$\mathbf{0.5t =2\ or\ t = 12}$

This gives

$\mathbf{t =4\ or\ t = 12}$

12 > 4.

Hence, the car comes to a complete stop at 12 seconds

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