Unformatted text preview: Clicker Quiz The average acceleration is defined as
A. B. C. D. Displacement change per time interval Position change per time interval Velocity change per time interval Speed change per time interval Physics for Scientists & Engineers 1
Spring Semester 2011
Lecture 4 Acceleration and Kinematic Equations January 19, 2011 Physics for Scientists&Engineers 1 1 January 19, 2011 Physics for Scientists&Engineers 1 2 Acceleration Vector
The average acceleration is defined as the velocity change per time interval ∆v ax = x ∆t The instantaneous acceleration is given by
ax = lim ax = lim
∆t → 0 ∆t → 0 Average/Instantaneous Acceleration Average/Instantaneous ∆vx dvx ≡ ∆t dt The acceleration vector is given by
a= dv dt
Average acceleration over Average acceleration over a large time interval a smaller time interval Instantaneous acceleration at time t3 January 19, 2011 Physics for Scientists&Engineers 1 3 January 19, 2011 Physics for Scientists&Engineers 1 4 Example: Velocity & Acceleration
Graph of x (t ) = 17.2 m − (10.1 m)(t / s)+(1.1 m)(t /s) and
v(t ) = −10.1 m/s+(2.2 m/s)(t / s)
2 Acceleration Concepts
We don’t have a separate word for the absolute value of the acceleration We refer to the deceleration of an object as a decrease in the speed the object over time, which corresponds to acceleration in the opposite direction of the motion of the object If the velocity and acceleration are in the same direction, the object moves faster If the velocity and acceleration are in opposite directions, the object slows down Acceleration:
January 19, 2011 a(t ) = 2.2 m/s 2
Physics for Scientists&Engineers 1 5 January 19, 2011 Physics for Scientists&Engineers 1 6 1 Clicker Quiz
When you’re driving a car along a straight road, you could be traveling in the positive or negative direction and you could have a positive acceleration or a negative acceleration. If you have negative velocity and positive acceleration you are A. B. C. D. Slowing down in the positive direction Speeding up in the negative direction Speeding up in the positive direction Slowing down in the negative direction Example: 100 m Sprint
Carl Lewis’ World Record, 1991 World Championship January 19, 2011 Physics for Scientists&Engineers 1 7 January 19, 2011 Physics for Scientists&Engineers 1 8 Example: 100 m Sprint (2)
vx = ∆x ∆t ax = ∆v x ∆t Displacement and Velocity from Acceleration Integration is the inverse operation to differentiation
• Fundamental Theorem of Calculus Fit: v = 11.6 m/s Fit: a = 0.0 m/s We can reverse the process we used to get from displacement to velocity to acceleration Why? We will obtain five kinematic equations that govern all of onedimensional motion
• assuming constant acceleration We won’t have to work them out every time we want to solve a problem
January 19, 2011 Physics for Scientists&Engineers 1 9 January 19, 2011 Physics for Scientists&Engineers 1 10 Displacement from Velocity
Start with the xcomponent of the velocity
vx ( t ) = dx ( t ) ⇒ dt t t dx ( t ') ∫ vx ( t ') dt ' = t∫ dt dt ' =x ( t ) − x ( t0 ) ⇒ t0 0 x ( t ) = x ( t0 ) + ∫ vx ( t ') dt ' ⇒
t0 t Velocity from Acceleration
Start with the xcomponent of the acceleration
ax ( t ) = dvx ( t ) ⇒ dt t t dvx ( t ') ∫ ax ( t ') dt ' = ∫ dt dt ' =vx ( t ) − vx ( t0 ) ⇒ t0 t0
t ax (t ) vx ( t ) = vx ( t0 ) + ∫ ax ( t ') dt ' ⇒
t0 x = x0 + ∫ vx ( t ') dt '
t0 t vx ( t ) = vx 0 + ∫ ax ( t ') dt '
t0 t January 19, 2011 Physics for Scientists&Engineers 1 11 January 19, 2011 Physics for Scientists&Engineers 1 12 2 Constant Acceleration
Many physical situations involve constant acceleration
t Constant Acceleration
Take constant ax
vx = vx 0 + ∫ ax dt ' = vx 0 + axt
t0 t vx = vx 0 + ∫ ax dt ' = vx 0 + axt
t0 1 x = x0 + ∫ vx dt ' = x0 + ∫ ( vx 0 + ax t ') dt ' = x0 + vx 0t + axt2 2 t0 t0 t t 1 x = x0 + ∫ vx dt ' = x0 + ∫ ( vx 0 + ax t ') dt ' = x0 + vx 0t + axt2 2 t0 t0 t t Now we have two of our five kinematic equations Let’s get the other three
January 19, 2011 Physics for Scientists&Engineers 1 13 January 19, 2011 Physics for Scientists&Engineers 1 14 Average Velocity for Constant Acceleration a Definition of average velocity
vx =
t Combine previous equations
vx = Average Velocity (2)
1 vx 0t + at 2 1 2 = vx 0 + at t 2 ∫ vx ( t ') dt '
0 ∫ v ( t ') dt '
x 0 t ∫ dt '
Denominator is easy
0 t ∫ dt '
0 t = ∫ dt ' = t
0 t Make an informed rearrangement 1 1 1 1 1 1 vx = vx 0 + at = vx 0 + vx 0 + at = vx 0 + ( vx 0 + at ) 2 2 2 2 2 2 And the final answer is 1 1 1 1 1 vx = vx 0 + ( vx 0 + at ) = vx 0 + vx = ( vx 0 + vx ) 2 2 2 2 2 So the average velocity is half the sum of the initial velocity plus the final velocity
15 January 19, 2011 Physics for Scientists&Engineers 1 16 Numerator ∫ v ( t ') dt ' = ∫ ( v
x 0 0
January 19, 2011 t t x0 1 + at ') dt ' = vx 0t + at2 2
Physics for Scientists&Engineers 1 More Kinematic Equations
The fourth kinematic equation is just
x = x0 + vxt The fifth and last kinematic equation requires some algebra Last Kinematic Equation
ax ( x − x0 ) = vx vx 0 − v2x 0 + 12 ( v x − 2vxvx 0 + v2x 0 ) = 1 v2x − 1 v2x0 2 2 2 So multiply by 2 and rearrange Start with vx = vx 0 + axt ⇒ t= vx − vx 0 ax
2 v2x = v2x 0 + 2ax ( x − x0 ) 1 v −v 1 v −v x = x0 + vx 0t + axt2 = x0 + vx 0 x x 0 + ax x x 0 2 ax 2 ax x = x0 + vx vx 0 − v2x 0 1 v2x − 2vx vx 0 + v2x 0 + ax 2 ax Now we subtract x0 and multiply by ax January 19, 2011 Physics for Scientists&Engineers 1 17 January 19, 2011 Physics for Scientists&Engineers 1 18 3 Our Five Kinematic Equations Example: Airplane Takeoff (1) TakeExperiment: use strain gauge and measure acceleration during airplane takeoff Result: Constant acceleration, in good approximation 1 (i) x = x0 + vx 0t + axt 2 2 (ii) x = x0 + vxt (iii) vx = vx 0 + axt 1 (iv) vx = ( vx + vx 0 ) 2 2 2 (v) vx = vx 0 + 2ax ( x − x0 )
January 19, 2011 Physics for Scientists&Engineers 1 19 a = 4.3 m/s2 January 19, 2011 Physics for Scientists&Engineers 1 20 Example: Airplane Takeoff (2) TakeQuestion 1:
Assuming a constant acceleration of a = 4.3 m/s2, starting from rest, what is the takeoff speed of the airplane reached after 18 seconds? Example: Airplane Takeoff (3) TakeQuestion 2:
How far down the runway has this airplane moved by the time it takes off? Answer 1:
The airplane accelerates from a standing start: initial 1 velocity is 0 x = x + v t + a t2
0 x0 x Answer 2:
x0 = 0 v0 = 0 a = 4.3 m/s 2 t1 = 18 s v(t ) = v0 + at x(t ) = x0 + v0t + 1 at 2 2 a = 4.3 m/s v0 = 0 t1 = 18 s 2 ←→ v(t1)? 2 x = x0 + vxt vx = vx 0 + axt ←x(t1)?→ v0→t = 1 [v0 + v(t )] 2 x(t ) = x0 + v0→ t t
2 v(t ) 2 = v0 + 2a[ x (t ) − x0 ] v(t1 ) = v0 + at1 = 0 + (4.3 m/s 2 )(18 s) = 77.4 m/s (175 mph)
January 19, 2011 Physics for Scientists&Engineers 1 21 January 19, 2011 1 ( vx + vx 0 ) 2 v = v + 2ax ( x − x0 ) vx =
2 x 2 x0 x(t1 ) = x0 + v0t1 + 1 at12 = 1 (4.3 m/s 2 )(18 s)2 2 2 = 697 m (2,290 ft)
Main runway at Lansing airport is 8506 ft long
22 Physics for Scientists&Engineers 1 Example: Top Fuel Racing (1) Example: Top Fuel Racing (2)
Accelerating from rest, a top fuel racer can reach 333.2 mph (=148.9 m/s, record established in 2003) at the end of a quarter mile (=402.3 m) run. For this example, we will assume constant acceleration. Question 1:
What is the value of this acceleration? Answer 1:
2 2 It is convenient in this case to use v (t ) = v0 + 2a[ x (t ) − x0 ] and solve for the acceleration a=
January 19, 2011 Physics for Scientists&Engineers 1 23 January 19, 2011 2 v(t ) 2 − v0 (148.9 m/s) 2 = = 27.6 m/s 2 2[ x(t ) − x0 ] 2(402.3 m) Physics for Scientists&Engineers 1 24 4 Example: Top Fuel Racing (4)
Question 2:
How long does it take to complete quarter mile race from a standing start? Answer 2:
Because the final velocity is 148.9 m/s, the average velocity is vx = 1 (148.9 m/s + 0) = 74.45 m/s 2 Now we can use
x(t ) = x0 + vxt ⇒ t = x(t ) − x0 402.3 m = = 5.40 s vx 74.45 m/s Real answer, measured on the track is t ≈ 4.5 s
Assumption of constant acceleration not quite satisfied
January 19, 2011 Physics for Scientists&Engineers 1 25 5 ...
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 Spring '08
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 Acceleration, Velocity, Five Kinematic Equations

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