a(t) = dv(t)dt dv(t) dt = 20 − 10t m/s2
v(1 s) = 15 m/s, v(2 s) = 20 m/s, v(3 s) = 15 m/s, v(5 s) = −25 m/s
a(1 s) = 10m/s2, a(2 s) = 0m/s2, a(3 s) = −10m/s2, a(5 s) = −30m/s2
At t = 1 s, velocity v(1 s) = 15 m/s is positive and acceleration is positive, so both velocity and acceleration are in the same direction. The particle is moving faster.
At t = 2 s, velocity has increased to v(2 s) = 20 m/s , where it is maximum, which corresponds to the time when the acceleration is zero. We see that the maximum velocity occurs when the slope of the velocity function is zero, which is just the zero of the acceleration function.
At t = 3 s, velocity is v(3 s) = 15 m/s and acceleration is negative. The particle has reduced its velocity and the acceleration vector is negative. The particle is slowing down.
At t = 5 s, velocity is v(5 s) = −25 m/s and acceleration is increasingly negative. Between the times t = 3 s and t = 5 s the particle has decreased its velocity to zero and then become negative, thus reversing its direction. The particle is now speeding up again, but in the opposite direction.
We can see these results graphically in Figure 3.4.7 .
Graph A shows velocity in meters per second plotted versus time in seconds. Velocity starts at zero, increases to 15 at 1 second, and reaches maximum of 20 at 2 seconds. It decreases to 15 at 3 seconds and continues to decrease to -25 at 5 seconds. Graph B shows acceleration in meters per second squared plotted versus time in seconds. Graph is linear and has a negative constant slope. Acceleration starts at 20 when time is zero, decreases to 10 at 1 second, to zero at 2 seconds, to -10 at 3 seconds, and to -30 and 5 seconds.
Graph A shows velocity in meters per second plotted versus time in seconds. Velocity starts at zero, increases to 15 at 1 second, and reaches maximum of 20 at 2 seconds. It decreases to 15 at 3 seconds and continues to decrease to -25 at 5 seconds. Graph B shows acceleration in meters per second squared plotted versus time in seconds. Graph is linear and has a negative constant slope. Acceleration starts at 20 when time is zero, decreases to 10 at 1 second, to zero at 2 seconds, to -10 at 3 seconds, and to -30 and 5 seconds.
Figure 3.4.7 : (a) Velocity versus time. Tangent lines are indicated at times 1, 2, and 3 s. The slopes of the tangents lines are the accelerations. At t = 3 s, velocity is positive. At t = 5 s, velocity is negative, indicating the particle has reversed direction. (b) Acceleration versus time. Comparing the values of accelerations given by the black dots with the corresponding slopes of the tangent lines (slopes of lines through black dots) in (a), we see they are identical.
Significance
By doing both a numerical and graphical analysis of velocity and acceleration of the particle, we can learn much about its motion. The numerical analysis complements the graphical analysis in giving a total view of the motion. The zero of the acceleration function corresponds to the maximum of the velocity in this example. Also in this example, when acceleration is positive and in the same direction as velocity, velocity increases. As acceleration tends toward zero, eventually becoming negative, the velocity reaches a maximum, after which it starts decreasing. If we wait long enough, velocity also becomes negative, indicating a reversal of direction. A real-world example of this type of motion is a car with a velocity that is increasing to a maximum, after which it starts slowing down, comes to a stop, then reverses direction.
EXERCISE 3.4
An airplane lands on a runway traveling east. Describe its acceleration.
Getting a Feel for Acceleration
You are probably used to experiencing acceleration when you step into an elevator, or step on the gas pedal in your car. However, acceleration is happening to many other objects in our universe with which we don’t have direct contact. Table 3.2 presents the acceleration of various objects. We can see the magnitudes of the accelerations extend over many orders of magnitude.
Explanation:
