# Acceleration

## A. Acceleration

### 1. Acceleration

When a particle’s velocity changes, the particle is said to undergo acceleration (or to accelerate). For motion along an axis, the average acceleration aavg over a time interval  $$\Delta t$$ is

$${{a}_{avg}}=\frac{{{v}_{2}}-{{v}_{1}}}{{{t}_{2}}-{{t}_{1}}}=\frac{\Delta v}{\Delta t}$$

Where the particle has velocity v1 at time t1 and then velocity v2 at time t2. The instantaneous acceleration (or simply acceleration) is   $$a=\frac{dv}{dt}$$     (2-8)

In words, the acceleration of a particle at any instant is the rate at which its velocity is changing at that instant. Graphically, the acceleration at any point is the slope of the curve of v(t) at that point. We can combine Eq.2-8 with Eq.2-4 to write

$$a=\frac{dv}{dt}=\frac{d}{dt}\left( \frac{dx}{dt} \right)=\frac{{{d}^{2}}x}{d{{t}^{2}}}$$   (2-9)

In world, the acceleration of a particle at any instant is the second derivative of its position x(t) with respect to time.

A common unit of acceleration of a particle at any instant is the second derivative of its position x(t) with respect to time.

A common unit of acceleration is the meter per second per second:  $$m/(s\cdot s)$$ or  $$m/{{s}^{2}}$$. Other units are in the form of length/(time $$\cdot$$ time) or length/time2. Acceleration has both magnitude and direction (it is yet another vector quantity). Its algebraic sign represents its direction on an axis just as for displacement and velocity; that is, acceleration with a positive value is in the positive direction of an axis, and acceleration with a negative value is in the negative direction.

Figure 2-6 (a) The x(t) curve for an elevator cab that moves upward along an x axis. (b) The v(t) curve for the cab. Note that it is the derivative of the x(t) curve ( $$v=\frac{dx}{dt}$$). (c) The a(t) curve for the cab. It it the derivative of the v(t) curve ( $$a=\frac{dv}{dt}$$). The stick figures along them bottom suggest how a passenger’s body might fell during the accelerations.

Figure 2-6 gives plots of the position, velocity, and acceleration of an elevator moving up a shaft. Compare the a(t) curve with the v(t) curve – each point on the a(t) curve shows the derivative (slope) of the v(t) curve at the corresponding time. When v is constant (at either 0 or 4 m/s), the derivative is zero and so also is the acceleration. When the cab first begins to move, the v(t)curve has a position derivative (the  slope is positive), which means that a(t) is positive. When the cab slows to a stop, the derivative and slope of the v(t) curve are negative; that is, a(t) is negative.

Next compare the slopes of the v(t) curve during the two acceleration periods. The slope associated with the cab’s slowing down (commonly called “deceleration”) is steeper because the cab stops in half the time it took to get up to speed. The steeper slope means that the magnitude of the deceleration is larger than that of the acceleration, as indicated in Fig.2-6c.

Sensations. The sensations you would feel while riding in the cab of Fig.2-6 are indicated by the sketched figures at the bottom. When the cab first accelerates, you feel as though you are pressed downward; when later the cab is braked to stop, you seem to be stretched upward. In between, you feel nothing special. In other words, your body reacts to accelerations (it is an accelerometer) but not to velocities (it is not a speedometer). When you are in a car traveling at 90 km/h or airplane traveling at 900 km/h, you have no bodily awareness of the motion. However, if the car or plane quickly changes velocity, you may become keenly aware of the change, perhaps even frightened by it.  Part of the thrill of an amusement park ride is due to the quick changes of velocity that you undergo (you pay for the accelerations, not for the speed). A more extreme example is shown in the photographs of Fig.2-7, which were taken while a rocket sled was rapidly accelerated along a track and then rapidly braked to stop.

Figure 2-7 Colonel J.P.Stapp in a rocket sled as it is brought up to higt speed (acceleration out of the page) and then very rapidly braked (acceleration into the page).

g Units. Large accelerations are sometimes expressed in terms of g units, with.

1 g = 9.8 m/s2 (g unit)   (2-10)

(As we shall discuss in Module 2-5, g is the magnitude of the acceleration of a falling object near Earth’s surface.) On a rooler coaster, you may experience brief accelerations up to 3g, which is (3)(9.8 m/s2), or about 29 m/s2, more than enough to justify the cost of the ride.

Signs. In common language, the sign of an acceleration has a nonscientific meaning: positive acceleration means that the speed of a object is increasing, and negative acceleration means that the speed is decreasing (the object is decelerating). In this book, however, the sign of an acceleration indicates a direction, not whether an object’s speed is increasing or decreasing. For example, if a car with an initial velocity  $$v=-25\text{ }m/s$$ is braked to a stop in 5.0 s, then  $${{a}_{avg}}=+5.0\text{ }m/{{s}^{2}}$$. The acceleration is positive, but the car’s speed has decreased. The reason is the difference in signs: the direction of the acceleration is opposite that of the velocity.

Here then is the proper way to interpret the signs: If the signs of the velocity and acceleration of a particle are the same, the speed of the particle increases. If the signs are opposite, the speed decreases.

### 2. Sample problem – Acceleration and  $$\frac{dv}{dt}$$

A particle’s position on the x axis of Fig.2-1 is given by  $$x=4-27t+{{t}^{3}}$$, with x in meters and t in seconds.

a) Because position x depends on time t, the particle must be moving. Find the particle’s velocity function v(t) and acceleration function a(t).

Key Ideas

(1) To get the velocity function v(t), we differentiate the position function x(t) with respect to time.

(2) To get the acceleration function a(t), we differentiate the velocity function v(t) with respect to time.

Calculations: Differentiating the position function, we find  $$v=-27+3{{t}^{2}}$$   (Answer)

with v in meters per second. Differentiating the velocity function then gives us  $$a=+6t$$   (Answer)

with a in meter per second squared.

b) Is there ever a time when v = 0?

Calculation: Setting v(t) = 0 yields  $$0=-27+3{{t}^{2}}$$, which has the solution  $$t=\pm 3\text{ }s$$  (Answer)

Thus, the velocity is zero both 3 s before and 3 s after the clock reads 0.

c) Decribe the particle’s motion for $$t\ge 0$$.

Reasoning: We need to examine the expressions for x(t), v(t), and a(t).

At t = 0, the particle is at x(0)=+4 m and is moving with a velocity of  $$v(0)=-27\text{ }m/s$$ – that is, in the negative direction of the x axis. Its accelerations is a(0) = 0 because just then the particle’s velocity is not changing (Fig.2-8a).

For 0 < t < 3 s, the particle still has a negative velocity, so it continues to move in the negative direction. However, its acceleration is no longer 0 but is increasing and positive. Because the signs of the velocity and the acceleration are opposite, the particle must be slowing (Fig.2-8b).

Indeed, we already know that it stops momentarily at t = 3 s. Just then the particle is as far to the left of the origin in Fig.2-1 as it will ever get. Substituting t = 3 s into the expression for x(t), we find that the particle’s position just then is x = -50 m (Fig.2-8c). Its acceleration is still positive.

For t > 3 s, the particle moves to the right on the axis. Its acceleration remains positive and grows progressively larger in magnitude. The velocity is now positive, and it too grows progressively larger in magnitude (Fig.2-8d)

## B. Sample problems

Example 1. The position of a particle moving along an x axis is given by $$x=12{{t}^{2}}-2{{t}^{3}}$$, where x is in meters and t is seconds. Determine (a) the position, (b) the velocity, and (c) the acceleration of the particle at t = 3.0 s. (d) What is the maximum positive coordinate reached by the particle and (e) at what time is it reached? (f) What is the maximun positive velocity reached by the particle and (g) at what time is it reached? (h) What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)? (i) Determine the average velocity of the particle between t = 0 and t = 3 s?

Solution:

a) Taking derivatives of $$x(t)=12{{t}^{2}}-2{{t}^{3}}$$ we obtain the velocity and the acceleration functions:

$$v(t)=24t-6{{t}^{2}}$$ and  $$a(t)=24-12t$$

With length in meters and time in seconds. Plugging in the value t = 3 yields x(3) = 54 m.

b) Similarly, plugging in the value t = 3 yields v(3) = 18 m/s.

c) For t = 3, $$a(3)=-12\text{ }m/{{s}^{2}}$$.

d) At the maximum x, we must have v = 0; eliminating the t = 0 root, the velocity equation reveals $$t=\frac{24}{6}=4\text{ }s$$ for the time of maximum x. Plugging t = 4 into the equation for x leads to x = 64 m for the largest x value reached by the particle.

e) From (d), we see that the x reaches its maximum at t = 4.0 s.

f) A maximum v requires a = 0, which occurs when $$t=\frac{24}{12}=2.0\text{ }s$$. This, inserted intro the velocity equation, gives vmax = 24 m/s.

g) From f), we see that the maximum of v occurs at $$t=\frac{24}{12}=2.0\text{ }s$$

h) In part e), the particle was (momentarily) motionless at t = 4 s. The acceleration at that time is readily found to be $$24-12(4)=-24\text{ }m/{{s}^{2}}$$.

i) The average velocity is defined by Eq.2-2, so we see that the values of x at t = 0 and t = 3 s are needed; these are, respectively, x = 0 and x = 54 m (found in part (a)).

Thus,  $${{v}_{avg}}=\frac{54-0}{3-0}=18\text{ }m/s$$.

Example 2. At a certain time a particle had a speed of 18 m/s in the positive x direction, and 2.4s later its speed was 30 m/s in the opposite direction. What is the average acceleration of the particle during this 2.4 s interval?

Solution:

Think. In this one-dimensional kinematics problems, we’re given the speed of a particle at two instants and asked to calculate its average acceleration.

Express. We represent the initial direction of motion as the +x direction. The average acceleration over a time interval  $${{t}_{1}}\le t\le {{t}_{2}}$$ is given by Eq.2-7:

$${{a}_{avg}}=\frac{\Delta v}{\Delta t}=\frac{v({{t}_{2}})-v({{t}_{1}})}{{{t}_{2}}-{{t}_{1}}}$$

Anlyze. Let  $${{v}_{1}}=+18\text{ }m/s$$ at t1 = 0 and  $${{v}_{2}}=-30\text{ }m/s$$ at t2 = 2.4 s. Using Eq.2-7 we find

$${{a}_{avg}}=\frac{v({{t}_{2}})-v({{t}_{1}})}{{{t}_{2}}-{{t}_{1}}}=\frac{-30\text{ }m/s-1\text{ }m/s}{2.4\text{ }s-0}=-20\text{ }m/{{s}^{2}}$$

Learn. The average acceleration has magnitude 20 m/s2 and is in the opposite direction to the particle’s initial velocity. This makes sense because the velocity of the particle is decreasing over the time interval. With t1 = 0, the velocity of the particle as a function of time can be written as

$$v={{v}_{0}}+at=(18\text{ }m/s)-(20\text{ }m/{{s}^{2}})t$$.

Example 3.

a) If the postion of a particle is given by $$x=20t-5{{t}^{3}}$$, where x is in meters and t is in seconds, when, if ever, is the particle’s velocity zero?

b) When is its acceleration a zero?

c) For what time range (positive or negative) is a negative?

d) Positive?

e) Graph x(t), v(t), and a(t).

Solution:

We use the functional natation x(t), v(t) and a(t) and find the latter two quantities by differentiating:

$$v(t)=\frac{dx(t)}{t}=-15{{t}^{2}}+20$$and  $$a(t)=\frac{dv(t)}{dt}=-30t$$

With SI units understood. These expressions are used in the parts that follow.

a) From $$0=-15{{t}^{2}}+20$$, we see that the only postitve value of t for which the particle is (momentarily) stopped is $$t=\sqrt{\frac{20}{15}}=1.2\text{ }s$$.

b) From $$0=-30t$$, we find a(0) = 0 (that is, it vanishes at t = 0).

c) It is clear that $$a(t)=-30t$$ is negative for t > 0.

d) The acceleration $$a(t)=-30t$$ is positive for t < 0.

e) The graphs are shown below. SI units are understood.

Example 4.

From t = 0 to t = 5.00 min, a man stands still, and from t = 5.00 min to t = 10.0 min, he walks briskly in a straight line at a constant speed of 2.20 m/s. What are (a) his average velocity vavg and (b) his average acceleration aavg in the time interval 2.00 min to 8.00 min? What are (c) vavg and (d) aavg in the time interval 3.00 min to 9.00 min? (e) Sketch x versus t and v versus t, and indicate how the answers to (a) through (d) can be obtained from the graphs.

Solution:

We use Eq.2-2 (average velocity) and Eq.2-7 (average acceleration). Regarding our coordinate choices, the initial position of the man is taken as the origin and his direction of motion during  $$5\min \le t\le 10\min$$ is taken to be the positive x direction. We also use the fact that  $$\Delta x=v\Delta {t}’$$ when the velocity is constant during a time interval  $$\Delta {t}’$$.

a) The entire interval considered is $$\Delta t=8-2=6\min$$, which is equivalent to 360 s, whereas the sub-interval in which he is moving is only  $$\Delta {t}’=8-5=3\min =180\text{ }s$$. His position at t = 2 min is x = 0 and his position at t = 8 min is $$x=v.\Delta {t}’=(2.2)(180)=396\text{ }m$$. Therefore,

$${{v}_{avg}}=\frac{396\text{ }m-0}{360\text{ }s}=1.10\text{ }m/s$$

b) The man is at rest at t = 2 min and has velocity v = +2.2 m/s at t = 8 min. Thus, keeping the answer to 3 signficant figures,

$${{a}_{avg}}=\frac{2.2\text{ }m/s-0}{360\text{ }s}=0.00611\text{ }m/{{s}^{2}}$$.

c) Now, the entire interval considered is $$\Delta t=9-3=6\min$$ (360 s again), whereas the sub-interval in which he is moving is  $$\Delta {t}’=9-5=4\min =240\text{ }s$$. His position at t = 3 min is x = 0 and his position t = 9 min is  $$x=v.\Delta {t}’=(2.2)(240)=528\text{ }m$$. Therefore,

$${{v}_{avg}}=\frac{\text{528 }m-0}{360\text{ }s}=1.47\text{ }m/s$$

d) The man is at rest at t = 3 min and has velocity v = +2.2 m/s at t = 9 min. Consequantly, $${{a}_{avg}}=\frac{2.2}{360}=0.00611\text{ }m/{{s}^{2}}$$ just as in part (b).

e) The horizontal line near the bottom of this x-vs-t graph represents the man standing at x = 0 for $$0\le t<300\text{ }s$$ and the linearly rising line for  $$300\le t\le 600\text{ }s$$ represents his constant-velocity motion. The lines represent the answers to part (a) and (c) in the sense that their slopes yield those results.

The graph of v-vs-t is not shown here, but would constist of two horizontal “stpes” (one at v = 0 for  $$0\le t<300\text{ }s$$ and the next at v = 2.2 m/s for  $$300\le t\le 600\text{ }s$$.) The indications of the average accelerations found in parts (b) and (d) would be dotted lines connecting the “steps” at the appropriate t values (the slopes of the dotted lines representing the values of aavg).

Example 5. The position of a particle moving along the x axis depends on the time according to the equation $$x=c{{t}^{2}}-b{{t}^{3}}$$, where x is in meters and t in seconds. What are the units of (a) constant c and (b) constant b? Let their numerical values be 3.0 and 2.0, respectively. (c) At what time does the particle reach its maximum positive x position? From t = 0.0 s to t = 4.0 s, (d) what distance does the particle move and (e) what is it displacement? Find its velocity at time (f) 1.0 s, (g) 2.0 s, (h) 3.0 s, and (i) 4.0 s. Find its acceleration at times (j) 1.0 s, (k) 2.0 s, (l) 3.0 s, and (m) 4.0 s.

Solution:

In this solution, we make use of the notation x(t) for the value of x at a particular t. The notations v(t) and a(t) have similar meanings.

a) Since the unit of ct2 is that of length, the unit of c must be that of length/time2, or m/s2 in the SI system.

b) Since bt3 has a unit of length, b must have a unit of length/time3, or m/s3.

c) When the particle reaches its maximum (or its minimum) coordinate its velocity is zero. Since the velocity is given by $$v=\frac{dx}{dt}=2ct-3b{{t}^{2}}$$, v = 0 occurs for t = 0 and for

$$t=\frac{2c}{3b}=\frac{2(3.0\text{ }m/{{s}^{2}})}{3(2.0\text{ }m/{{s}^{3}})}=1.0\text{ }s$$

For t = 0, x = xO = 0 and for t = 1.0 s, x = 1.0 m > xO. Since we seek the maximum, we reject the first root (t = 0) and accept the second (t = 1 s).

d) In the first 4 s the particle moves from the origin to x = 1.0 m, turns around, and goes back to

$$x(4\text{ }s)=(3.0\text{ }m/{{s}^{2}}){{(4.0\text{ }s)}^{2}}-(2.0\text{ }m/{{s}^{3}}){{(4.0\text{ }s)}^{3}}=-80\text{ }m$$

The total path length it travels is 1.0 m + 1.0 m + 80 m = 82 m.

e) Its displacement is $$\Delta x={{x}_{2}}-{{x}_{1}}$$, where x1 = 0 and $${{x}_{2}}=-80\text{ }m$$. Thus,  $$\Delta x=-80\text{ }m$$.

The velocity is given by  $$v=2ct-3b{{t}^{2}}=(6.0\text{ }m/{{s}^{2}})t-(6.0\text{ }m/{{s}^{3}}){{t}^{2}}$$

f) Plugging in t = 1 s, we obtain

$$v(1\text{ }s)=(6.0\text{ }m/{{s}^{2}})(1.0\text{ }s)-(6.0\text{ }m/{{s}^{3}}){{(1.0\text{ }s)}^{2}}=0$$.

g) Similarly, $$v(2\text{ }s)=(6.0\text{ }m/{{s}^{2}})(2.0\text{ }s)-(6.0\text{ }m/{{s}^{3}}){{(2.0\text{ }s)}^{2}}=-12\text{ }m/s$$.

h) $$v(3\text{ }s)=(6.0\text{ }m/{{s}^{2}})(3.0\text{ }s)-(6.0\text{ }m/{{s}^{3}}){{(3.0\text{ }s)}^{2}}=-36\text{ }m/s$$.

i) $$v(4\text{ }s)=(6.0\text{ }m/{{s}^{2}})(4.0\text{ }s)-(6.0\text{ }m/{{s}^{3}}){{(4.0\text{ }s)}^{2}}=-72\text{ }m/s$$

The acceleration is given by  $$a=\frac{dv}{dt}=2c-6b=6.0m/{{s}^{2}}-(12.0\text{ }m/{{s}^{3}})t$$.

j) Plugging in t = 1 s, we obtain $$a(1\text{ }s)=6.0\text{ }m/{{s}^{2}}-(12.0\text{ }m/{{s}^{3}})(1.0\text{ }s)=-6.0\text{ }m/{{s}^{2}}$$.

k) $$a(2\text{ }s)=6.0\text{ }m/{{s}^{2}}-(12.0\text{ }m/{{s}^{3}})(2.0\text{ }s)=-18\text{ }m/{{s}^{2}}$$ .

l) $$a(3\text{ }s)=6.0\text{ }m/{{s}^{2}}-(12.0\text{ }m/{{s}^{3}})(3.0\text{ }s)=-30\text{ }m/{{s}^{2}}$$.

m) $$a(4\text{ }s)=6.0\text{ }m/{{s}^{2}}-(12.0\text{ }m/{{s}^{3}})(4.0\text{ }s)=-42\text{ }m/{{s}^{2}}$$.

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