Motion With Constant Acceleration Young Freedman

2.4 Motion with Constant Acceleration • The simplest accelerated motion is straight-line motion with constant acceleration. • For this case, velocity changes at the same rate throughout the motion. • Such motions occur frequently in nature and also in human technology. • We will derive key equations for straight-line motion with constant acceleration.

2.4 Motion with Constant Acceleration • The motion diagram shows the position, velocity, and acceleration at five different times for a particle moving with constant acceleration.

2.4 Motion with Constant Acceleration • Since acceleration ax is constant, the ax-t graph shows a horizontal line.

• The graph of velocity vs. time has a constant slope because the acceleration is constant.

2.4 Motion with Constant Acceleration • We replace average acceleration aav-x by the constant (instantaneous) ax, v2 x − v1x ax = (2.7) t2 − t1 • Now we let t1 = 0 and t2 be any arbitrary later time t. We use υ0x for the x-component at the later time t is υx. Then Eqn 2.7 becomes vx − v0 x ax = or t −0 υ x = v0 x + a xt ( 2.8)

2.4 Motion with Constant Acceleration • Next, we derive an equation for the position of x of a particle moving with constant acceleration. • Firstly, we use Eqn 2.2, the initial position where time t = 0 and denoted by x0. • The position at the later time t is simply x. Thus for time interval ∆t = t − 0 and ∆x = x − x0, x − x0 ( 2.9) υav − x = t

2.4 Motion with Constant Acceleration • We can also obtain a second expression for υav-x that is valid for constant acceleration only. υ0 x − υ0 ( 2.10) υav − x = 2

• Substitute Eqn 2.8 into above equation,

υav − x

1 = ( υ0 x + υ0 x + a x t ) 2 1 = υ0 x + a x t 2

( 2.11)

2.4 Motion with Constant Acceleration • Finally, equating Eqns 2.9 and 2.11 and simplifying the result: 1 2 x = x0 + υ0 xt + a xt 2

( 2.12)

• Just as change in velocity of particle equals the area under ax-t graph, the displacement equals the area under νx-t graph. • This method is always valid even if the acceleration is not constant.

2.4 Motion with Constant Acceleration • Taking derivative of Eqn 2.12: dx υ x = = υ0 x + a xt dt • Differentiating again, dυ x = ax dt • In many problems, it is useful to have a relationship between position, velocity, and acceleration that does not include time.

2.4 Motion with Constant Acceleration • Solving Eqn 2,8 for t, then substitute resulting expression into Eqn 2.12, υ x − υ0 x t= ax  υ x − υ0 x  1  υ x − υ0 x  x = x0 + υ0 x   + a x    ax  2  ax 

2

2a x ( x − x0 ) = 2υ0 xυ x − 2υ0 x + υ x − 2υ0 xυ x + υ0 x 2

2

• Simplifying,

υ x = υ0 x + 2a x ( x − x0 ) 2

2

(2.13)

2

2.4 Motion with Constant Acceleration • Equating Eqns 2.9 and 2.10, multiply through by t,

υ0 x + υ x   x − x0 =  t 2  

(2.14)

• Eqns 2.8, 2.12, 2.13 and 2.14 are the equations of motion with constant acceleration. • These equations can be used to solve ANY kinematics problems involving straight-line motion of a particle with constant acceleration.

2.4 Motion with Constant Acceleration • Graph shows the coordinate x as a function of time for motion with constant acceleration. • Characteristics: – A parabola – Intercept x0 at t = 0 – Slope of tangent at t = 0 equals ν0x – Slope of tangent at t equals νx

Motion with Constant Acceleration

a

dv  dv  adt. If we integrate both sides of the equation we get: dt

 dv   adt  a  dt  v  at  C.

Here C is the integration constant.

C can be determined if we know the velocity v0  v(0) at t  0: v(0)  v0  (a )(0)  C  C  v0  v  v0  at

(eq. 1)

dx  dx  vdt   v0  at  dt  v0 dt  atdt. If we integrate both sides we get: dt at 2  dx   v0 dt a  tdt  x  v0t  2  C . Here C  is the integration constant. C  can be determined if we know the position xo  x(0) at t  0: v

a x(0)  xo  (v0 )(0)  (0)  C   C   xo 2 at 2 x(t )  xo  v0t  (eq. 2) 2

v  v0  at

(eq. 1) ;

If we eliminate the time t v 2  v02  2a  x  x0 

at 2 x  x0  v0t  (eq. 2) 2 between equation 1 and equation 2 we get:

(eq. 3)

Below we plot the position x(t ), the velocity v(t ), and the acceleration a versus time t: at 2 x  x0  v0t  2 The x(t) versus t plot is a parabola that intercepts the vertical axis at x = x0. v  v0  at The v(t) versus t plot is a straight line with slope = a and intercept = v0. The acceleration a is a constant. (2-9)

2.4 Motion with Constant Acceleration 1.2 (Analyzing Motion Using Graphs) 1.3 (Analyzing Motion from Graphs) 1.4 (Predicting Motion from Graphs) 1.5 (Problem-Solving Strategies for Kinematics) 1.6 (Skier Races Downhill) 1.8 (Seat Belts Save Lives) Applet 1.9 (Screeching to a Halt) 1.11 (Car Starts, then Stops) 1.12 (Solving Two-Vehicle Problems) 1.13 (Car Catches Truck) 1.14 (Avoiding a Rear-End Collision)

2.4 Motion with Constant Acceleration Problem-solving strategy (Motion with constant acceleration) • IDENTIFY: • In most straight-line motion problems, you can use the constant-acceleration equations. • Occasionally, however, you will encounter a situation in which the acceleration isn’t constant. In such a case, you’ll need a different approach (see Section 2.6).

2.4 Motion with Constant Acceleration Problem-solving strategy (Motion with constant acceleration) • SET UP: • You must decide at the beginning of a problem where the origin of coordinates is and which axis direction is positive. These choices are usually a matter of convenience. It is easiest to place the particle at the origin at time t = 0; then x0 = 0. It is always helpful to make a motion diagram showing these choices and some later positions of the particle.

2.4 Motion with Constant Acceleration Problem-solving strategy (Motion with constant acceleration) • SET UP: • Remember that your choice of the positive axis direction automatically determines the positive directions for velocity and acceleration. If x is positive to the right of the origin, then νx and ax are also positive toward the right.

2.4 Motion with Constant Acceleration Problem-solving strategy (Motion with constant acceleration) • SET UP: 1. Restate the problem in words first, then translate this description into symbols and equations. When does the particle arrive at a certain point (what is the value of t)? Where is the particle when its velocity has a specified value (what is the value of x when νx has the specified value)?

2.4 Motion with Constant Acceleration Problem-solving strategy (Motion with constant acceleration) • SET UP: • Make a list of quantities such as x, x0, νx, ax, and t. In general, some of them will be known and some will be unknown. Write down the values of the known quantities, and decide which of the unknowns are the target variables. Be on the lookout for implicit information.

2.4 Motion with Constant Acceleration Problem-solving strategy (Motion with constant acceleration) • EXECUTE: 1. Choose an equation from Eqns. 2.8, 2.12, 2.13, and 2.14 that contains only one of the target variables. 2. Solve this equation for the target variable, using symbols only. 3. Substitute the known values and compute the value of the target variable. 4. Sometimes you will have to solve 2 simultaneous equations for 2 unknown quantities.

2.4 Motion with Constant Acceleration Problem-solving strategy (Motion with constant acceleration) • EVALUATE: 1. Take a hard look at your results to see whether they make sense. Are they within the general range of values you expected?

Example 2.4 Constant-acceleration calculations A motorcyclist heading east through a small Iowa city accelerates after he passes the signpost marking the city limits. His acceleration is a constant 4.0 m/s2. At time t = 0 he is 5.0 m east of the signpost, moving east at 15 m/s. (a) Find his position and velocity at time t = 2.0 s. (b) Where is the motorcyclist when his velocity is 25 m/s?

Example 2.4 (SOLN) Identify: The statement of the problem tells us explicitly that the acceleration is constant, so we can use the constantacceleration equations. Set Up: We take the signpost as the origin of coordinates (x = 0), and choose the positive x-axis to point east. At the initial time t = 0, the initial position is x0 = 5.0 m and the initial velocity is ν0x = 15 m/s. The constant acceleration is ax = 4.0 m/s2. The unknown target variables in part (a) are the values of the position x and the velocity νx at the later time t = 2.0 s; the target variable in part (b) is the value of x when νx = 25 m/s.

Example 2.4 (SOLN) Execute: (a) We can find the position at t = 2.0 s by using Eqn 2.12, which gives position x as a function of time t: 1 2 x = x0 + υ0 xt + a xt 2 1 = 5.0 m + (15 m/s )( 2.0 s ) + 4.0 m/s 2 ( 2.0 s ) 2 2 = 43 m We can find the velocity at this same time by using Eqn 2.8, which gives velocity νx as a function of time υ x = υ0 x + a xt t:

(

(

)

)

= 15 m/s + 4.0 m/s2 ( 2.0 s ) = 23 m/s

Example 2.4 (SOLN) Execute: (b) From our soln to part (a), we see that the velocity is νx = 25 m/s at a time later than 2.0 s and at a point farther than 43 m from the signpost. From Eqn. 2.13 we have υ 2 = υ 2 + 2a ( x − x ) x

0x

x

0

Solving for x and substituting in the known values, we 2 2 υ x − υ0 x find x = x0 + 2a x = 5.0 m + = 55 m

( 25 m/s ) 2 − (15 m/s ) 2

(

2 4.0 m/s 2

)

Example 2.4 (SOLN) Execute: (b) Alternatively, we may use Eqn 2.8 to first find the time when νx = 25 m/s: υ x = υ0 x + a xt so υ x − υ0 x 25 m/s − 15 m/s t= = ax 4.0 m/s 2

t = 2.5 s Then from Eqn. 2.12, we have 1 2 x = x0 + υ0 xt + a xt 2

(

)

1 = 5.0 m + ( 15 m/s )( 2.5 s ) + 4.0 m/s 2 ( 2.5 s ) 2 2 = 55 m

Example 2.4 (SOLN) Evaluate: Do these results make sense? According to our results in part (a), the motorcyclist accelerates from 15 m/s (about 34 mi/h, or 54 km/h) to 23 m/s (about 51 mi/h, or 83 km/h) in 2.0 s while traveling a distance of 38 m (about 125 ft). Our results in part (b) tell us that after an additional 0.5 s, the motorcyclist has moved an additional 12 m (about 39 ft) and has accelerated to 25 m/s (56 mi/h, or 90 km/h). This is pretty brisk acceleration, but well within the capabilities of a highperformance bike.

Example 2.5 Two bodies with different accelerations

A motorist traveling with constant velocity of 15 m/s (about 34 mi/h) passes a school-crossing comer, where the speed limit is 10 m/s (about 22 mi/h). Just as the motorist passes, a police officer on a motorcycle stopped at the comer starts off in pursuit with constant acceleration of 3.0 m/s2. (a) How much time elapses before the officer catches up with the motorist? (b) What is the officer's speed at that point? (c) What is the total distance each vehicle has traveled at that point?

Example 2.5 (SOLN)

Identify: The police officer and the motorist both move with constant acceleration (equal to zero for the motorist), so we can use the formulas we have developed. Set Up: We take the origin at the corner, so x0 = 0 for both, and take the positive direction to the right. Let xP (for police) be the officer's position and xM (for motorist) be the motorist's position at any time. The initial velocities are νP0x = 0 for the officer and νM0x = 15 m/s for the motorist; the constant accelerations are aPx = 3.0 m/s2 for the officer and aMx = 0 for the motorist.

Example 2.5 (SOLN) Set Up: Our target variable in part (a) is the time when the officer catches the motorist; when the two vehicles are at the same position. In part (b) we're looking for the officer's speed ν (the magnitude of his velocity) at the time found in part (a). In part (c) we want to find the position of either vehicle at this same time. Hence we use Eqn. 2.12 in parts (a) and (c), and Eqn. 2.8 in part (b). Execute: (a) We want to find the value of the time t when the motorist and the police officer are at the same position: xM = xP.

Example 2.5 (SOLN)

Execute: (a) Applying Eqn. 2.12 to each vehicle, we get 1 2 xM = 0 + υM0 xt + ( 0 ) t = υM0 xt 2 1 2 1 xP = 0 + ( 0 ) t + ( aPx ) t = ( aPx ) t 2 2 2 Since xM = xP at time t, we set these two expressions equal to each other and solve for t: 1 υM0 xt = aPxt 2 2 2υM0 x 2(15 m/s ) t=0 or t= = = 10 s 2 aPx 3.0 m/s

Example 2.5 (SOLN)

Execute: (a) There are two times when both the vehicles have the same x-coordinate. The first, t = 0, is the time when the motorist passes the parked motorcycle at the corner. The second, t = 10 s, is the time when the officer catches up with the motorist. (b) We want the magnitude of the officer's velocity νPx at the time t found in part (a). Her velocity at any time is given by Eqn. 2.8: υPx = υP0 x + aPxt = 0 + 3.0 m/s2 t

(

so when t = 10 s, we find νPx = 30 m/s.

)

Example 2.5 (SOLN)

Execute: (b) When the officer overtakes the motorist, she is traveling twice as fast as the motorist is. (c) In 10 s the distance the motorist travels is xM = υM0 xt = (15 m/s )(10 s ) = 150 m and the distance the officer travels is 1 2 1 xP = aPxt = 3.0 m/s 2 (10 s ) 2 = 150 m 2 2

(

)

This verifies that at the time the officer catches the motorist, they have gone equal distances.

Example 2.5 (SOLN)

Evaluate: Graphs of x vs. t for each vehicle shown. We see that there are 2 times when the 2 positions are the same. At neither of these times do the 2 vehicles have the same velocity. At t = 0, the officer is at rest; at t = 10 s, the officer has twice the speed of the motorist. In real pursuits the officer would accelerate to a speed faster than that of the motorist, then slow down to have the same velocity as the motorist when she catches him. We haven't treated this case here because it involves a changing acceleration.