Lecture 3 Motion at constant acceleration an important special case An example of motion at constant acceleration is motion near the earth s surface due to gravity To restrict this motion to one dimension we consider just up and down motion Later when we study motion in two dimensions we will analyse projectile motion eg shoot the monkey It is possible to write down formulas for displacement and velocity when the acceleration is constant To do this first notice that if the acceleration a is constant then the velocity increases at a constant rate This means that if we plot velocity as a function of time it is a straight line with slope a We thus deduce that the velocity as a function of time has the form v v0 at 1 Now the average velocity over a time period t is given by x 1 v v0 v 2 t 2 Here we used Eq 2 of Lecture 2 but used t instead of t We also dropped the subscipts used in Lecture 2 This makes the formula more compact but we have to keep in mind the meaning of each of the variables Using Eq 1 for v in Eq 2 we find 1 x v0 v0 at 2 t 3 and solving for x leads to 1 x v0 t at2 2 4 We may also eliminate t from this equation in favor of v using Eq 1 after some algebra this yields v 2 v02 2a x 5 Eqs 1 2 4 and 5 are the key constant acceleration formula and they appear in many problems due to their broad importance These formula will also be the basis of understanding many types of motion in two dimensions 1 An application Whooping it up An excited cowboy fires his handgun at head height vertically upward with an initial speed v0 311m s Ignoring air drag on the bullet how long before the bullet lands on the cowboy s head How high does the bullet go Take the gravitational acceleration to be 9 81m s2 Note that this acceleration is the same regardless of whether the motion is positive up or negative down To find the time taken for the bullet to reach its highest altitude note that at the highest point of its motion v 0 so we must have using Eq 1 0 311m s 9 81m s2 t 6 which gives t 31 7s The time until the bullet hits the cowboy s head is then 2t 63 4s The displacement at the top of its trajectory is found from Eq 4 which yields x 4930m about 3 miles What is the velocity of the bullet when it hits the cowboy Answer 311m s How would these results be affected by air drag 2
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