2.7 Advanced Topic: Instantaneous Acceleration

Although the constant-acceleration model is very useful, real moving objects only rarely have constant acceleration. For example, FIGURE 2.36a is a realistic velocity-versus-time graph for a car leaving a stop sign. The graph is not a straight line, so this is not motion with constant acceleration.

The figure illustrates two graphs labeled ("a") and (b) that depict the velocity- and acceleration-versus-time graphs of a car leaving a stop sign.

FIGURE 2.36 Full Alternative Text

We can define an instantaneous acceleration much as we defined the instantaneous velocity. The instantaneous velocity at time $t$ is the slope of the position-versus-time graph at that time or, mathematically, the derivative of the position with respect to time. By analogy: The instantaneous acceleration $a_{s}$ is the slope of the line that is tangent to the velocity-versus-time curve at time $t$ . Mathematically, this is

a_{s} = \frac{d v_{s}}{d t} = slope of the velocity-versus-time graph at time t

(2.27)

FIGURE 2.36b applies this idea by showing the car’s acceleration graph. At each instant of time, the value of the car’s acceleration is the slope of its velocity graph. The initially steep slope indicates a large initial acceleration. The acceleration decreases to zero as the car reaches cruising speed.

The reverse problem—to find the velocity $v_{s}$ if we know the acceleration $a_{s}$ at all instants of time—is also important. Again, with analogy to velocity and position, we have

v_{f s} = v_{i s} + \int_{t_{i}}^{t_{f}} a_{s} d t

(2.28)

The graphical interpretation of Equation 2.28 is

v_{f s} = v_{i s} + area under the acceleration curve a_{s} between t_{i} a n d t_{f}

(2.29)

EXAMPLE 2.17 Finding velocity from acceleration

FIGURE 2.37 shows the acceleration graph for a particle with an initial velocity of $10 m / s$ . What is the particle’s velocity at $t = 8$ s?

The figure illustrates an acceleration-versus-time graph of a particle.

FIGURE 2.37 Full Alternative Text

MODEL We’re told this is the motion of a particle.

VISUALIZE FIGURE 2.37 is a graphical representation of the motion.

SOLVE The change in velocity is found as the area under the acceleration curve:

v_{f s} = v_{i s} + area under the acceleration curve a_{s} between t_{i} a n d t_{f}

The area under the curve between $t_{i} = 0 s$ and $t_{f} = 8 s$ can be subdivided into a rectangle $(0 s \leq t \leq 4 s)$ and a triangle $(4 s \leq t \leq 8 s)$ . These areas are easily computed. Thus

\begin{array}{l} v_{s} (at t = 8 s) & = & 10 m / s + (4 m / s^{2}) (4 s) \\ + \frac{1}{2} (4 m / s^{2}) (4 s) \\ = & 34 m / s \end{array}

EXAMPLE 2.18 A realistic car acceleration

Starting from rest, a car takes $T$ seconds to reach its cruising speed $v_{max}$ . A plausible expression for the velocity as a function of time is

v_{x} (t) = {\begin{matrix} v_{max} (\frac{2 t}{T} - \frac{t^{2}}{T^{2}}) & t \leq T \\ v_{max} & t \geq T \end{matrix}

Demonstrate that this is a plausible function by drawing velocity and acceleration graphs.
Find an expression for the distance traveled at time $T$ in terms of $T$ and the maximum acceleration $a_{max}$ .
What are the maximum acceleration and the distance traveled for a car that reaches a cruising speed of $15 m / s$ in 8.0 s?

MODEL Model the car as a particle.

VISUALIZE FIGURE 2.38a shows the velocity graph. It’s an inverted parabola that reaches $v_{max}$ at time $T$ and then holds that value. From the slope, we see that the acceleration should start at a maximum value $a_{max}$ , steadily decrease until $T$ , and be zero for $t > T$ .

The figure consists of two parts labeled ("a") and (b) that depict a velocity-versus-time graph and an acceleration-versus-time graph of a car.

FIGURE 2.38 Full Alternative Text

SOLVE a. We can find an expression for $a_{x}$ by taking the derivative of $v_{x}$ . Starting with $t \leq T$ , and using Equation 2.6 for the derivatives of polynomials, we find

a_{x} = \frac{d v_{x}}{d t} = v_{max} (\frac{2}{T} - \frac{2 t}{T^{2}}) = \frac{2 v_{max}}{T} (1 - \frac{t}{T}) = a_{max} (1 - \frac{t}{T})

where $a_{max} = 2 v_{max} / T$ . For $t \geq T$ , $a_{x} = 0$ . Altogether,

a_{x} (t) = {\begin{array}{l} a_{max} (1 - \frac{t}{T}) & t \leq T \\ 0 & t \geq T \end{array}

This expression for the acceleration is graphed in FIGURE 2.38b. The acceleration decreases linearly from $a_{max}$ to 0 as the car accelerates from rest to its cruising speed.

b. To find the position as a function of time, we need to integrate the velocity (Equation 2.11) using Equation 2.13 for the integrals of polynomials. At time $T$ , when cruising speed is reached,

\begin{array}{l} x_{T} & = x_{0} + \int_{0}^{T} v_{x} d t = 0 + \frac{2 v_{\max}}{T} \int_{0}^{T} t d t - \frac{v_{\max}}{T^{2}} \int_{0}^{T} t^{2} d t \\ = \frac{2 v_{\max}}{T} \frac{t^{2}}{2} |_{0}^{T} - \frac{v_{\max}}{T^{2}} \frac{t^{3}}{3} |_{0}^{T} \\ = v_{\max} T - \frac{1}{3} v_{\max} T = \frac{2}{3} v_{\max} T \end{array}

Recalling that $a_{max} = 2 v_{max} / T$ , we can write the distance traveled as

x_{T} = \frac{2}{3} v_{max} T = \frac{1}{3} (\frac{2 v_{max}}{T}) T^{2} = \frac{1}{3} a_{max} T^{2}

If the acceleration stayed constant, the distance would be $\frac{1}{2} a T^{2}$ . We have found a similar expression but, because the acceleration is steadily decreasing, with a smaller fraction in front.

c. With $v_{max} = 15 m / s$ and $T = 8.0 s,$ realistic values for city driving, we find

\begin{array}{l} a_{max} = \frac{2 v_{max}}{T} = \frac{2 (15 m/s)}{8.0 s} = 3.75 {m/s}^{2} \\ x_{T} = \frac{1}{3} a_{max} T^{2} = \frac{1}{3} (3.75 {m/s}^{2}) {(8.0 s)}^{2} = 80 m \end{array}

REVIEW 80 m in 8.0 s to reach a cruising speed of 15 $m / s \approx$ 30 mph is very reasonable. This gives us good reason to believe that a car’s initial acceleration is $\approx \frac{1}{3} g$ .

CHAPTER 2 CHALLENGE EXAMPLE Rocketing along

A rocket sled accelerates along a long, horizontal rail. Starting from rest, two rockets burn for 10 s, providing a constant acceleration. One rocket then burns out, halving the acceleration, but the other burns for an additional 5 s to boost the sled’s speed to 625 m/s. How far has the sled traveled when the second rocket burns out?

MODEL Model the rocket sled as a particle with constant acceleration.

VISUALIZE FIGURE 2.39 shows the pictorial representation. This is a two-part problem with a beginning, an end (the second rocket burns out), and a point in between where the motion changes (the first rocket burns out).

A pictorial representation of the motion of a rocket sled.

FIGURE 2.39 Full Alternative Text

SOLVE The difficulty with this problem is that there’s not enough information to completely analyze either the first or the second part of the motion. A successful solution will require combining information about both parts of the motion, and that can be done only by working algebraically, not worrying about numbers until the end of the problem. A well-drawn pictorial representation and clearly defined symbols are essential.

The first part of the motion, with both rockets firing, has acceleration $a_{0 x}$ . The sled’s position and velocity when the first rocket burns out are

\begin{array}{l} x_{1} = x_{0} + v_{0 x} Δ t + \frac{1}{2} a_{0 x} {(Δ t)}^{2} = \frac{1}{2} a_{0 x} t_{1}^{2} \\ v_{1 x} = v_{0 x} + a_{0 x} Δ t = a_{0 x} t_{1} \end{array}

where we simplified as much as possible by knowing that the sled started from rest at the origin at $t_{0} = 0 s$ . We can’t compute numerical values, but these are valid algebraic expressions that we can carry over to the second part of the motion.

From $t_{1}$ to $t_{2}$ , the acceleration is a smaller $a_{1 x}$ . The velocity when the second rocket burns out is

v_{2 x} = v_{1 x} + a_{1 x} Δ t = a_{0 x} t_{1} + a_{1 x} (t_{2} - t_{1})

where for $v_{1 x}$ we used the algebraic result from the first part of the motion. Now we have enough information to complete the solution. We know that the acceleration is halved when the first rocket burns out, so $a_{1 x} = \frac{1}{2} a_{0 x}$ . Thus

v_{2 x} = 625 m/s = a_{0 x} (10 s) + \frac{1}{2} a_{0 x} (5 s) = (12.5 s) a_{0 x}

Solving, we find $a_{0 x} = 50 {m/s}^{2}$ .

With the acceleration now known, we can calculate the position and velocity when the first rocket burns out:

\begin{matrix} x_{1} & = & \frac{1}{2} a_{0 x} t_{1}^{2} = \frac{1}{2} (50 {m/s}^{2}) {(10 s)}^{2} = 2500 m \\ v_{1 x} & = & a_{0 x} t_{1} = (50 {m/s}^{2}) (10 s) = 500 m/s \end{matrix}

Finally, the position when the second rocket burns out is

\begin{matrix} x_{2} & = & x_{1} + v_{1 x} Δ t + \frac{1}{2} a_{1 x} {(Δ t)}^{2} \\ = & 2500 m + (500 m/s) (5 s) + \frac{1}{2} (25 {m/s}^{2}) (5 {s)}^{2} = 5300 m \end{matrix}

The sled has traveled 5300 m when it reaches $625 m / s$ at the burnout of the second rocket.

REVIEW 5300 m is 5.3 km, or roughly 3 miles. That’s a long way to travel in 15 s! But the sled reaches incredibly high speeds. At the final speed of 625 m/s, over 1200 mph, the sled would travel nearly 10 km in 15 s. So 5.3 km in 15 s for the accelerating sled seems reasonable.

2.7 Advanced Topic: Instantaneous Acceleration

FIGURE 2.36 Velocity and acceleration graphs of a car leaving a stop sign.

EXAMPLE 2.17 Finding velocity from acceleration

FIGURE 2.37 Acceleration graph for Example 2.17.

EXAMPLE 2.18 A realistic car acceleration

FIGURE 2.38 Velocity and acceleration graphs for Example 2.18.

CHAPTER 2 CHALLENGE EXAMPLE Rocketing along

FIGURE 2.39 The pictorial representation of the rocket sled.