# Thread: I'm not sure how to solve this question?

1. ## I'm not sure how to solve this question?

Question: At t = 55 seconds, an object moving along a straight path is 47 meters from its origin. If the velocity at that time is v(55) = −3 meters per second, use a tangent line approximation to predict the objects position at one minute (t = 60 seconds).

So I know that the equation for a tangent line is:
T(x) = f(x) + f'(x) (x-c)

But I am not sure if that is how I solve this question as I'm trying to predict the object's position.

Any help is appreciated.

2. Let $\displaystyle p(t)$ be the position in meters of the object at time $\displaystyle t$ seconds. Then, $\displaystyle p(t) = -4.9t^2 + v_0t + s_0$ where $\displaystyle v_0$ is the initial velocity in meters per second of the object and $\displaystyle s_0$ is the initial position in meters of the object.

Let $\displaystyle L(t)$ be the tangent line approximation to p(t) at t = 55. Then, $\displaystyle L(t) = p(55) + p'(55)(t - 55)$. Because $\displaystyle p(55) = 47$ and $\displaystyle p'(55) = v(55) = -3$, $\displaystyle L(t) = 47 - 3(t - 55)$.

Evaluate $\displaystyle L(60)$.

3. Thank you for answering!
So to evaluate L(60)..
L(60) = p(60) + p'(60)(x - 60)
Is that correct?

4. Sorry, I had let $\displaystyle L(t) = p(55) + p'(55)(x - 55)$ in my original post. The variable $\displaystyle x$ shouldn't be on the left-hand side of the equation.

I meant to let $\displaystyle L(t) = p(55) + p'(55)(t - 55)$.

Then, $\displaystyle L(60) = p(55) + p'(55)(60 - 55) = 47 - 3(5) = 47 - 15 = 32$.

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