1. Linear Kinematics - displacement, velocity and acceleration Contents:
Quandary
New formula
Deriving the rest of the formulas
How to solve these: suvat
Whiteboards

2. Quandary
A car goes from 14 m/s to 26 m/s in 10. seconds. How far does it go in this time?
it averages 20 m/s for 10 seconds, so it goes 200 m
TOC

3. A new formula: s = (u + v)t 2 s - displacement (m)
u - initial velocity (m/s)
v - final velocity (m/s)
a - acceleration (m/s/s)
t - time (s)
s = (u + v)t 2
A car goes from 14 m/s to 26 m/s in 10. seconds. How far does it go in this time?
s = (14 m/s + 26 m/s)(10. s) = 200 m = 2.0 x 102 m 2
A person starts from rest and goes 14.5 m in 3.75 s. What is their final velocity?
14.5 m = (0 + v)(3.75 s), v = = 7.73 m/s 2
TOC

4. Deriving: - 2 more nifty formulas!!
v = u + at
s = (u + v)t 2
From the first formula:
t = (v - u) a
Derive:
v2 = u2 + 2as
s = ut + 1/2at2
when to use which one…
s - displacement (m)
u - initial velocity (m/s)
v - final velocity (m/s)
a - acceleration (m/s/s)
t - time (s)
TOC

6. Example #1 A car goes from 14 m/s to 26 m/s in 300. m.
What is the acceleration, and
What time does it take?
Show suvat
use v2 = u2 + 2as, and s = (u+v)t/2
TOC

7. Example #2
A rocket going 3130 m/s accelerates at m/s/s for a distance of 5.50 x 109 m.
What time does it take, and
What is the final velocity?
suvat
s = ut + 1/2at2 is hard
v2 = u2 + 2as can be done
(then use v = u + at)
TOC

8. Whiteboards:
suvat
1 | 2 | 3 | 4
TOC

9. A cart stops in a distance of 3. 81 m in a time of 4. 51 s
A cart stops in a distance of 3.81 m in a time of 4.51 s. What was its initial velocity?
s = 3.81 m, u = ??, v = 0, t = 4.51 s
s = (u + v)t 2
u = m/s, 1.69 m/s W
1.69 m/s

10. A car going 12 m/s accelerates at 1. 2 m/s/s for 5. 0 seconds
A car going 12 m/s accelerates at 1.2 m/s/s for 5.0 seconds. What is its displacement during this time?
s = ?, u = 12 m/s, v = ??, a = 1.2 m/s/s, t = 5.0 s
s = ut + 1/2at2
s =75 m, 75 m W
75 m

11. Another car with a velocity of 27 m/s stops in a distance of 36. 74 m
Another car with a velocity of 27 m/s stops in a distance of m. What was its acceleration?
s = m, u = 27 m/s, v = 0, a = ??, t = ??
v2 = u2 + 2as
a = m/s/s, -9.9 m/s/s W
-9.9 m/s/s

12. A car’s brakes slow it at 9. 5 m/s/s. If it stops in 47
A car’s brakes slow it at 9.5 m/s/s. If it stops in 47.3 m, how fast was it going to start with?
s = 47.3 m, u = ??, v = 0, a = -9.5 m/s/s, t = ??
v2 = u2 + 2as
u = m/s, 30. m/s W
30. m/s

13. Another different car has a final velocity of 12
Another different car has a final velocity of 12.0 m/s after having decelerated for a distance of 78.0 m in 4.50 seconds. What was the acceleration of the car?
(Hint – use no a for u, then no s for a)
solution W
m/s/s

14. What time will it take a car going 23 m/s to start with, and accelerating at 3.5 m/s/s, to go 450 m?
s = 450 m, u = 23 m/s, v = ??, a = 3.5 m/s/s, t = ??
s = ut + 1/2at2
(Oh Nooooooo!)
OR - use v2 = u2 + 2as - find v, and use v = u + at
t = s, 11 s W
11 s

15. Start
Basic Problem Solving
Is it
Bigger than
Your
Mouth?
Yes
Cut it up a bit No
Eat it
Done

16. Start
Basic Problem Solving
Write down given quantities
Are
you done
yet? No
Apply any formula
Yes
Circle the answer

17. What distance does a train go if it slows from 32. 0 m/s to 5
What distance does a train go if it slows from 32.0 m/s to 5.0 m/s in 135 seconds?
s = ?, u = 32.0 m/s, v = 5.0 m/s, t = 135 s
s = (u + v)t 2
s = m, 2.50x103 m W
2.50x103 m

18. A ship slows from 18 m/s to 12 m/s over a distance of 312 m
A ship slows from 18 m/s to 12 m/s over a distance of 312 m. What time did it take?
s = 312 m, u = 18 m/s, v = 12 m/s, t = ??
s = (u + v)t 2
t =20.8 s, 21 s W
21 s