This is an impressive displacement to cover in only 5.56 s, but top-notch dragsters can do a quarter mile in even less time than this. If we convert 402 m to miles, we find that the distance covered is very close to one-quarter of a mile, the standard distance for drag racing. We can combine the previous equations to find a third equation that allows us to calculate the final position of an object experiencing constant acceleration. Solving for Final Position with Constant Acceleration Note that it is always useful to examine basic equations in light of our intuition and experience to check that they do indeed describe nature accurately. If a is negative, then the final velocity is less than the initial velocityĪll these observations fit our intuition.If the acceleration is zero, then the final velocity equals the initial velocity ( v = v 0), as expected (in other words, velocity is constant).Final velocity depends on how large the acceleration is and how long it lasts.In addition to being useful in problem solving, the equation v = v 0 + a t v = v 0 + a t gives us insight into the relationships among velocity, acceleration, and time. With jet engines, reverse thrust can be maintained long enough to stop the plane and start moving it backward, which is indicated by a negative final velocity, but is not the case here. The final velocity is much less than the initial velocity, as desired when slowing down, but is still positive (see figure). Note the acceleration is negative because its direction is opposite to its velocity, which is positive. To summarize, using the simplified notation, with the initial time taken to be zero,įigure 3.19 The airplane lands with an initial velocity of 70.0 m/s and slows to a final velocity of 10.0 m/s before heading for the terminal. Also, it simplifies the expression for change in velocity, which is now Δ v = v − v 0 Δ v = v − v 0. It also simplifies the expression for x displacement, which is now Δ x = x − x 0 Δ x = x − x 0. This gives a simpler expression for elapsed time, Δ t = t Δ t = t. That is, t is the final time, x is the final position, and v is the final velocity. We put no subscripts on the final values.
That is, x 0 x 0 is the initial position and v 0 v 0 is the initial velocity. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity. Since elapsed time is Δ t = t f − t 0 Δ t = t f − t 0, taking t 0 = 0 t 0 = 0 means that Δ t = t f Δ t = t f, the final time on the stopwatch.
Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification.
Notationįirst, let us make some simplifications in notation. Then we investigate the motion of two objects, called two-body pursuit problems. We first investigate a single object in motion, called single-body motion. In this section, we look at some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration. But, we have not developed a specific equation that relates acceleration and displacement. You might guess that the greater the acceleration of, say, a car moving away from a stop sign, the greater the car’s displacement in a given time.
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