# The acceleration for an object moving on a line (given in m/s^2) is given by a(t) = 2t-2. Its...

## Question:

The acceleration for an object moving on a line (given in {eq}m/s^2 {/eq}) is given by {eq}a(t) = 2t-2 {/eq}. Its initial velocity is {eq}-8m / s {/eq} (that is , {eq}v(0) = -8 {/eq}).

(a) Find the displacement of the object on the time interval {eq}0 \leq t \leq 5. {/eq}

(b) Find the total distance traveled on the time interval {eq}0 \leq t \leq 5. {/eq}

## Acceleration, Velocity and Distance:

Integration is used to obtain function from one another. Acceleration is the rate of change of velocity i.e. acceleration is equal to the derivative of velocity with respect to time. So velocity is the integral of acceleration with respect to time and distance is the integral of velocity with respect to time.

## Answer and Explanation:

{eq}a(t)=2t-2 {/eq}

Integrate to obtain the velocity function

{eq}v(t)=\int a(t)dt \\ v(t)= \int (2t-2)dt \\ v(t)=t^2-2t+C {/eq}

It is given that {eq}v(0)= -8 {/eq}

{eq}v(t)=t^2-2t-8 {/eq}

A). Displacement {eq}r(t)=\int _{a}^{b} v(t)dt {/eq}

Velocity {eq}v(t)=t^2-2t-8 {/eq}

As range of time {eq}0 \leq t \leq 5 {/eq}

So limits of integration are {eq}a=0, b=5 {/eq}

Thus {eq}r(t)=\int _{0}^{5}(t^2-2t-8)dt {/eq}

Solving integration we get

{eq}r(t)= \left | \frac{t^{3}}{3}-t^{2}-8t \right |_{0}^{5} \\ r(t)= [ \frac{125}{3}-65 ] \\ r(t)= -\frac{70}{3} {/eq}

{eq}B). \ Distance \ is \ the \ absolute \ value. Hence \\ d=\int \left | v(t)\right | dt \\ d=\int _{0}^{5} | t^2-2t-8 | dt {/eq}

As velocity is negative on the interval {eq}[0,4] {/eq} and positive on the interval {eq}[4,5] {/eq}

{eq}Hence \ d=\int _{0}^{4} (t^2-2t-8) dt + \int _{4}^{5} (t^2-2t-8) dt \\ d=\int _{0}^{4} (8+2t-t^2) dt + \int _{4}^{5} (t^2-2t-8) dt {/eq}

Solving integral we get, {eq}d= \left | 8t+t^2-\frac{t^3}{3} \right |_{0}^{4}+\left | \frac{t^3}{3}-t^2-8t \right |_{4}^{5} \\ d=\frac{80}{3}+10 \\ d=\frac{110}{3} {/eq}

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from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13