1. Accelerated Motion Problems
1. An automobile manufacturer claims that its sports car can accelerate from rest to a speed of 39.0 m/s in 8.0 seconds.
Determine the acceleration of the car
a = vf –vi/Δt = 39.0 m/s/8.0 seconds = 4.9 m/s2
b) Find the distance the car would travel in those 8 seconds
X = ½ at2 + vit = ½ (4.9m/s2)(8s)2
= 157 m

2. Problem #2
A racing car starting from rest accelerates at a rate of 4.5 m/s2. What is the velocity after it has traveled for 10 seconds?
Vf = at + vi = 4.5 m/s2 (10s) = m/s + 0

3. Problem #3
A train running at 30 m/s is slowed uniformly to a stop in 44 seconds. Find the acceleration and stopping distance.
a = vf – vi/t = 0 – 30 m/s / 44 seconds = m/s2
x = ½ at2 + vit = ½ (-.68 m/s2)(44s) m/s (44s)
=-658 m m = 662 m

4. Problem #4
A car is traveling at 60 km/h. It accelerates at 3m/s2 for 5 seconds. What speed does it reach?
60 km/h x 1000m/1km x 1h/3600 s = 16.7 m/s
vf = at + vi = 3m/s2 (5s) m/s = 31.7 m/s

5. Problem #5
An object falls freely from rest. Find the distance it falls in 3.5 seconds. Find the final speed it reaches after falling this time.
a = 9.8 m/s2 x = ½ at2 + vit
x = ½ (9.8 m/s2)(3.5s)2 = 60 m

6. Acceleration Problems set #2
There are four basic equations for accelerated motion problems. If you identify the given information, you can apply the appropriate equation
vf = at + vi x = ½ at2 + vit
x = ½ (vi + vf) ax = vf2 – vi2

7. Acceleration problems set #2
An airplane must have a velocity of 50 m/s in order to take off. What must be the airplane’s acceleration if the runway is 500 meters long?
vf = 50 m/s x= 500 m vi = 0 a=?
2ax = vf2 –vi2 a = vf2 –vi2/2x
a = 2.5 m/s2

8. Acceleration problems set #2
A car traveling 10 m/s accelerates to a speed of 25 m/s in 12 seconds. What distance does it cover in that time?
vi = 10 m/s vf = 25 m/s t = 12 seconds
x=?
X = ½( vf + vi)t = ½( 10m/s + 25m/s) 12 seconds
x = 210 meters

9. Acceleration problems set #2
A car moving at 35 m/s is brought to a stop in 12 seconds. What is the acceleration? What distance does it cover?
vf=0 vi = 35 m/s t = 12 s
a = vf – vi/t = -35 m/s/12 s
a = -2.9 m/s2
x = ½(vf+vi)t = 210 meters

10. Projectile motion
Key Idea – The horizontal motion is independent of the vertical motion.

11. Projectile Motion
Horizontally launched projectile. The vertical motion is identical to free fall

12. Projectile Motion
Example: A motorcycle rider jumps off a cliff. His horizontal motion is described by:
X=vxt
His vertical motion is described by Y = 1/2gt2

13. Projectile motion
Example: A motorcycle rider jumps off a 35 meter high cliff at 45 m/s. How far from the edge of the cliff does he land?

14. Projectile motion Example #2
Example: An airplane moving at a constant velocity of 115 meters/second and altitude of 1050 meters drops a package. How far horizontally does it fall from its point of release?

15. Projectile motion Example #3
For the package released from the airplane, what is its speed just before it hits the ground?

16. Projectile Motion Projectiles launched at an angle:
The initial velocity must first be resolved into vertical and horizontal components
Vy = vsinθ
Vx = vcosθ

17. Projectile Motion
Example#4 A placekicker kicks a football at an angle of 40 degrees with a speed of 22 m/s. Find the time of flight, range, and maximum height.
Vx = vcosθ = 22m/s(cos40) = 16.9 m/s
Vy = vsinθ = 22m/s(sin40) = 14.1 m/s
T (time to highest point) = vy/g = 1.4 seconds
Time of flight = 2T = 2.8 seconds
Range = vx 2T = 16.9 m/s (2.8 seconds)
= 47.3 meters
Maximum height Y = 1/2gt2 + vyt
=-1/2 (9.8m/s2)(1.4s) m/s(1.4 s)
= 10.1 meters

18. Home Run?
A ball player hits a pitch 30m/s at 20o, does it clear a fence 5 meters high, 90 meters away? First resolve the velocity into x and y components
vx = vcos20o = 28.2 m/s
Vy = vsin20o = 10.3 m/s
The time we need to find is the time to the fence tf
tf = 90m/ 28.2 m/s = 3.2 seconds
How high is the ball at that time?
Y =-1/2 gt2 +vyt = m + 33 m = m
The ball never reaches the fence.