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  5. Motion with constant acceleration

3.4 Motion with constant acceleration  (Page 5/10)

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Top figure shows cars located at 64.3 meters and 90 meters from the starting point for dry and wet conditions, respectively. Bottom figure shows cars located at 79.3 meters and 105 meters from the starting point for dry and wet conditions, respectively.
The distance necessary to stop a car varies greatly, depending on road conditions and driver reaction time. Shown here are the braking distances for dry and wet pavement, as calculated in this example, for a car traveling initially at 30.0 m/s. Also shown are the total distances traveled from the point when the driver first sees a light turn red, assuming a 0.500-s reaction time.

Significance

The displacements found in this example seem reasonable for stopping a fast-moving car. It should take longer to stop a car on wet pavement than dry. It is interesting that reaction time adds significantly to the displacements, but more important is the general approach to solving problems. We identify the knowns and the quantities to be determined, then find an appropriate equation. If there is more than one unknown, we need as many independent equations as there are unknowns to solve. There is often more than one way to solve a problem. The various parts of this example can, in fact, be solved by other methods, but the solutions presented here are the shortest.

Calculating time

Suppose a car merges into freeway traffic on a 200-m-long ramp. If its initial velocity is 10.0 m/s and it accelerates at 2.00 m/s 2 , how long does it take the car to travel the 200 m up the ramp? (Such information might be useful to a traffic engineer.)

Strategy

First, we draw a sketch [link] . We are asked to solve for time t . As before, we identify the known quantities to choose a convenient physical relationship (that is, an equation with one unknown, t .)

Figure shows car accelerating from the speed of 10 meters per second at a rate of 2 meters per second squared. Acceleration distance is 200 meters.
Sketch of a car accelerating on a freeway ramp.

Solution

Again, we identify the knowns and what we want to solve for. We know that x 0 = 0 ,
v 0 = 10 m/s , a = 2.00 m/ s 2 , and x = 200 m.

We need to solve for t . The equation x = x 0 + v 0 t + 1 2 a t 2 works best because the only unknown in the equation is the variable t , for which we need to solve. From this insight we see that when we input the knowns into the equation, we end up with a quadratic equation.

We need to rearrange the equation to solve for t , then substituting the knowns into the equation:

200 m = 0 m + ( 10.0 m/s ) t + 1 2 ( 2.00 m/s 2 ) t 2 .

We then simplify the equation. The units of meters cancel because they are in each term. We can get the units of seconds to cancel by taking t = t s, where t is the magnitude of time and s is the unit. Doing so leaves

200 = 10 t + t 2 .

We then use the quadratic formula to solve for t ,

t 2 + 10 t − 200 = 0 t = − b ± b 2 − 4 a c 2 a ,

which yields two solutions: t = 10.0 and t = −20.0. A negative value for time is unreasonable, since it would mean the event happened 20 s before the motion began. We can discard that solution. Thus,

t = 10.0 s .

Significance

Whenever an equation contains an unknown squared, there are two solutions. In some problems both solutions are meaningful; in others, only one solution is reasonable. The 10.0-s answer seems reasonable for a typical freeway on-ramp.

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Check Your Understanding A manned rocket accelerates at a rate of 20 m/s 2 during launch. How long does it take the rocket to reach a velocity of 400 m/s?

To answer this, choose an equation that allows us to solve for time t , given only a , v 0 , and v :
v = v 0 + a t .
Rearrange to solve for t :
t = v − v 0 a = 400 m/s − 0 m/s 20 m/s 2 = 20 s .

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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