4.1.3 Constant Acceleration

If the acceleration a⃗=axî+ay ĵ is constant, then the two components ax and ay are both constant. In this case, everything you learned about constant-acceleration kinematics in 《 SECTION 2.4 carries over to two-dimensional motion.

Consider a particle that moves with constant acceleration from an initial position r⃗i=xi î+yi ĵ, starting with initial velocity v ⃗i=vix î+viy ĵ. Its position and velocity at a final point f are

xf=xi+vix Δt+12ax (Δt)2yf=yi+viy Δt+12ay (Δt)2vfx=vix+ax Δtvfy=viy+ay Δt (4.11)

There are many quantities to keep track of in two-dimensional kinematics, making the pictorial representation all the more important as a problem-solving tool.

NOTE   For constant acceleration, the x-component of the motion and the y-component of the motion are independent of each other. However, they remain connected through the fact that Δt must be the same for both.

EXAMPLE 4.3 Plotting a spacecraft trajectory

In the distant future, a small spacecraft is drifting “north” through the galaxy at 680 m/s when it receives a command to return to the starship. The pilot rotates the spacecraft until the nose is pointed 25° north of east, then engages the ion engine. The spacecraft accelerates at 75 m/s2. Plot the spacecraft’s trajectory for the first 20 s.

MODEL Model the spacecraft as a particle with constant acceleration.

VISUALIZE FIGURE 4.7 shows a pictorial representation in which the y-axis points north and the spacecraft starts at the origin. Notice that each point in the motion is labeled with two positions (x and y), two velocity components (vx and vy), and the time t. This will be our standard labeling scheme for trajectory problems.

FIGURE 4.7 Pictorial representation of the spacecraft.

The figure illustrates a pictorial representation of the trajectory of a spacecraft in the x y coordinate plane.

SOLVE The acceleration vector has both x- and y-components; their values have been calculated in the pictorial representation. But it is a constant acceleration, so we can write

x1=x0+v0x(t1−t0)+12 ax(t1−t0)2=34.0 t12 my1=y0+v0y(t1−t0)+12 ay(t1−t0)2=680t1+15.8t12 m

where t1 is in s. Graphing software produces the trajectory shown in FIGURE 4.8. The trajectory is a parabola, which is characteristic of two-dimensional motion with constant acceleration.

FIGURE 4.8 The spacecraft trajectory.

The figure illustrates a graph of the trajectory of a spacecraft on a coordinate plane.