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The initial velocity and acceleration of four moving objects at a given instant in time are given...

Question:

The initial velocity and acceleration of four moving objects at a given instant in time are given in the following table. Determine the final speed of each of the objects, assuming that the time elapsed since {eq}t = 0 \ s {/eq} is {eq}2.0 \ s. {/eq}

$$\text{Initial velocity}$$ $$\text{Acceleration} $$
$$+13 \ m/s $$ $$-3.3 \ m/s^2 $$
$$+13 \ m/s$$ $$-3.5 \ m/s^2 $$
$$-12 \ m/s$$ $$+3.1 \ m/s^2 $$
$$-13 \ m/s$$ $$-3.0 \ m/s^2$$

Uniform Acceleration:

Uniform acceleration motion, also known as constant acceleration is a type of motion in which the velocity of moving object changes at a constant rate. There are 3 equations of motion that are used to study objects moving with the constant acceleration. These equations are:

{eq}\displaystyle { v = u + at \\ s = ut+ 0.5at^2 \\ v^2-u^2=2as } {/eq}

The negative sign of acceleration means the velocity is decreasing with respect to time. This is known as deceleration. Equations of motion in case of deceleration can be written as:

{eq}\displaystyle { v = u + (-a)t = u -at \\ s = ut+ 0.5(-a)t^2 = ut-0.5at^2 \\ v^2-u^2=2(-a)s = -2as } {/eq}

Answer and Explanation:


Let the initial speed be u and the acceleration be a.


Object 1:

The final speed of the object 1 is,

{eq}v = u + at \\ = +13 + (-3.3)(2) \\ = 13-6.6 \\ = 6.4 \ m/s {/eq}


Object 2:

The final speed of the object 2 is,

{eq}v = u + at \\ = +13 + (-3.5)(2) \\ = 13-7 \\ = 6 \ m/s {/eq}


Object 3:

The final speed of the object 3 is,

{eq}v = u + at \\ = -12 + (3.1)(2) \\ = 13-6.2 \\ = 6.8 \ m/s {/eq}


Object 4:

The final speed of the object 4 is,

{eq}v = u + at \\ = -13 + (-3)(2) \\ = -13-6 \\ = -19 \ m/s {/eq}

which is the answer.


Learn more about this topic:

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Kinematic Equations List: Calculating Motion

from AP Physics 1: Exam Prep

Chapter 4 / Lesson 15
19K

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