Describe the velocity change 2
Q. Driver’s manual published by the highway departments of most states used to point out the dangers of fast driving by showing, usually graphically, the disproportionality of large stopping distances for high velocities. The figures below shows a typically chart.
Stopping Distance: From eye to brain to foot to wheel to road
25 MPH, Thinking distance= 27 ft; Braking Distance = 34.4 ft (61.4 ft)
35 MPH, Thinking Distance = 38 ft; Braking Distance = 67 ft. (105 ft)
45 MPH, Thinking Distance = 49 ft; Braking Distance = 110 ft. (159 ft.)
55 MPH, Thinking Distance = 60 ft; Braking Distance = 165 ft. (225 ft.)
65 MPH; Thinking Distance = 71 ft; Braking Distance = 231 ft. (302 ft.)
The figures converted to m and s.
|
| Velocity (m/s) | Thinking distance (m) | Braking distance (m) | Total |
| 1 | 11.18 | 8.23 | 10.49 | 18.72 |
| 2 | 15.65 | 11.58 | 20.42 | 32.0 |
| 3 | 20.12 | 14.94 | 33.53 | 48.47 |
| 4 | 24.59 | 18.29 | 50.29 | 68.58 |
| 5 | 29.06 | 21.64 | 70.41 | 92.05 |
a) What is the reaction time (in seconds) associated with the “thinking distance” for each of the initial speeds? Explain your reasoning and what physics principles did you use?
Reaction time = Thinking distance / speed
Physics principle: Speed remains constant during the reaction time associated with thinking distance.
|
| Thinking distance / speed | Reaction Time s |
| 1 | 8.23/10.49 | 0.736 |
| 2 | 11.58/15.65 | 0.74 |
| 3 | 14.94/20.12 | 0.742 |
| 4 | 18.29/24.59 | 0.744 |
| 5 | 21.64/29.06 | 0.745 |
b) Based on the data given in the chart, what is the acceleration of the car (include magnitude and direction) and is it independent of the initial velocity? Express your final result with the correct significant figures.
v2 - u2 = 2as
v = 0
therefore u2 = 2as
or a = -u2/2s
|
| -u2/2s | Acceleration (m/s2) |
| 1 | -(11.18)2/2x10.49 | -5.96 |
| 2 | -(15.65)2/2x20.42 | -6.0 |
| 3 | -(20.12)2/2x33.53 | -6.04 |
| 4 | -(24.59)2/2x50.29 | -6.01 |
| 5 | -(29.06)2/2x70.41 | -6.0 |
The acceleration is more or less the same for all initial velocities.
c) For each initial velocity, how long does it take the driver to stop from the time he/she begins to think about stopping? What physics principles did you use?
Braking time t2 :
v = u + at
v = 0
so t = -u/a
|
| -u/a | t2 (s) | Total time (s) |
| 1 | -11.8/-5.96 | 1.876 | 1.876 + 0.736 = 2.612 |
| 2 | -15.65/-6.00 | 2.608 | 2.608 + 0.74 = 3.348 |
| 3 | -20.12/-6.04 | 3.331 | 3.331 + 0.742 = 4.073 |
| 4 | -24.59/-6.01 | 4.092 | 4.092 + 0.744 = 4.836 |
| 5 | -29.06/-6 | 4.843 | 4.843 + 0.745 = 5.588 |
Q. A ball is thrown down vertically with an initial speed of 20.5 m/s from a height of 58.8 m.
a) What will be its speed just before it strikes the ground?
u = 20.5 m/s
h = 58.8 m
According to equation of motion
v2 - u2 = 2gh
so v = (u2 + 2gh)1/2
= [(20.5)2 + 2 x 9.8 x 58.8]1/2
= 39.7 m/s
b) How long will it take for the ball to reach the ground?
v = u + gt (equation of motion)
t = (v - u)/g = (39.7 - 20.5)/9.8
= 1.96 s
c) What would be the answers to a) and b) if the ball were thrown directly up from the same height and with the same initial speed?
v would be same that is 39.7 m/s
t = (39.7 - (-20.5))/9.8 = 6.14 s (initial velocity in the upward direction will be taken as negative)
d) How long will it take for the ball to reach maximum height?
At the maximum height
v = 0
u = -20.5 m/s
a = g = 9.8 m/s2
v = u + gt
0 = -20.5 + 9.8t
so t = 20.5/9.8 = 2.09 s
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