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Q Describe the velocity change: a. Suppose an object undergoes a change in velocity from +1m/s to 4m/s. Is its velocity becoming more positive or less positive than it was before? What direction (s) is the object moving? Is the acceleration positive or negative? Explain your response. Are the velocity and acceleration in the same direction during the time interval of the change in velocity? Explain. Initial velocity u = 1 m/s. Final velocity v = 4 m/s. Velocity is becoming more positive as v - u = 4 - 1 = 3 m/s is positive. Object is moving in the positive direction. Acceleration is positive as the velocity is increasing. Yes, the velocity and acceleration are in the same direction during the time interval of the change in velocity as only then the velocity can increase. We can see this from the first equation of motion: v = u + at v is greater than u only if a is positive that is in the same direction as v. b. Suppose an object undergoes a change in velocity from –4m/s to /1m/s. Is its velocity becoming more positive or less positive than it was before? What direction (s) is the object moving? Is the acceleration positive or negative? Are the velocity and acceleration in the same direction during the time interval of the change in velocity? Explain Velocity is becoming more positive u = - 4 m/s v = 1 m/s Change in velocity = v - u = 1 - (-4) = 5 m/s which is positive. Initially the object is moving in the negative direction. When the force is applied, it decelerates and continues to move in the negative direction. Eventually its velocity becomes zero and then it starts to move in the positive direction. Thus acceleration is positive. Initially the velocity and acceleration are in opposite directions till velocity becomes zero. Then acceleration and velocity are in the same direction This is because first the body decelerate, its velocity going from -4m/s to 0 m/s, then it changes direction, and body accelerates from 0 to 1 m/s c. Suppose an object is turning around so that it undergoes a change in velocity from –2m/s to +2m/s. Is its velocity becoming more positive or less positive than it was before? What direction (s) is the object moving? Is the acceleration positive or negative? Are the velocity and acceleration in the same direction during the time interval of the change in velocity? Explain. Velocity is becoming more positive. u = -2 m/s and v = 2 m/s v - u = change in velocity = 2 - (-2) = 4 m/s Object is moving in the negative direction initially. It decelerates, stops and begins to move in the positive direction. The acceleration is positive. Initially velocity is negative, acceleration is positive, that is why the body decelerates. When the velocity of object becomes zero, the velocity and acceleration start acting in the same direction and the body's speed increases from 0 to 2 m/s. d. An object is turning around so that it undergoes a change in velocity from +1 m/s to –1m/s. Is its velocity more positive or less positive than it was before? What direction (s) is the object moving? Is it undergoing acceleration while it is turning around? Is the acceleration positive or negative? Are the velocity and acceleration in the same direction during the time interval of the change in velocity? u = 1 m/s , v = - 1 m/s Change in velocity = v-u = -1 -1 = -2 m/s Velocity is less positive than it was before. The object is initially moving in the positive direction. It decelerates stops and starts moving in the negative direction. First it undergoes deceleration till speed becomes zero then it accelerates. Acceleration is negative. Velocity and acceleration are in the opposite direction when body is moving towards the positive direction. After it stops and changes direction, velocity and acceleration are in same direction that is both acting along the negative direction |
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
Q Two boys are standing on the top of a wall which is 400 m from the ground. The first boy drops a tennis ball over the side of the wall. Two seconds later, his friend throws another ball vertically downward. Calculate the required initial velocity of the second ball in order for it to hit the ground at the same time as the first ball.
