04 homework YOO, HEE Due: Jan 26 2008, 4:00 am Question 1, chap 2, sect 4. part 1 of 1 10 points A car is moving at constant speed on the freeway. The driver sees a patrol car at time t1 and rapidly slows down by around 10 miles per hour. After continuing at this speed for a few minutes, the driver at time t2 returns to the earlier constant speed. Let us plot the acceleration of the car as a function of time; take the forward direction of motion as positive. Which of the following graphs correctly describes the car's acceleration a(t)? 1. 6.
a 0 time a 0 0 t1 t2 0 t1 t2 time
1
a 0
time t1 t2
0
5.
a 0
time t1 t2
0
2. 7.
a 0 time a 0 0 t1 t2 0 t1 t2 time
3.
correct 8.
a 0 time a 0 0 t1 t2 0 t1 t2 time
4. Explanation: The car is at first moving at a constant speed and therefore its acceleration is zero. After seeing a patrol car it starts slowing down i.e. decelerating for a short while. During
homework 04 YOO, HEE Due: Jan 26 2008, 4:00 am that time period its acceleration is negative. The acceleration soon goes back to zero and stays at zero while the car moves at a constant (but lower) speed once again. At time t2 the car briefly accelerates (and so its acceleration is positive for a short while) to the original constant speed at which it remains (and so its acceleration goes back to zero). Question 2, chap 2, sect 4. part 1 of 1 10 points A record of travel along a straight path is as follows: (a) Start from rest with constant acceleration of 2.97 m/s2 for 15.4 s; (b) Constant velocity for the next 0.91 min; (c) Constant negative acceleration of -9.91 m/s2 for 5.61 s. What was the total displacement x for the complete trip? Correct answer: 2950.12 m (tolerance 1 %). Explanation: This trip is divided into three sections: 1 (a) Acceleration from rest: xa = at2 2 (b) Constant velocity motion: xb = vt 1 (c) Deceleration: x = vt + at2 2 Question 3, chap 2, sect 5. part 1 of 2 10 points An electron has an initial speed of 260000 m/s. If it undergoes an acceleration of 2.1 1014 m/s2 , how long will it take to reach a speed of 522000 m/s? Correct answer: 1.24762 10-9 s (tolerance 1 %). Explanation: Basic Concepts: v = v0 + a t 1 x = x0 + v0 t + a t2 2 Solution: Assuming the acceleration given is an average acceleration, we can use v = v0 +at Solution:
2 v 2 = v0 + 2axstop
2
and solve for t: t= v - v0 522000 m/s - 260000 m/s = a 2.1 1014 m/s2 = 1.24762 10-9 s . Question 4, chap 2, sect 5. part 2 of 2 10 points How far has it traveled in this time? Correct answer: 0.000487819 m (tolerance 1 %). Explanation: Plugging t into our formula for x gives us 1 x = x0 + v0 t + a t2 2 = 0.000487819 m . Question 5, chap 2, sect 5. part 1 of 2 10 points A motorist is traveling at 14 m/s when he sees a deer in the road 46 m ahead. If the maximum negative acceleration of the vehicle is -7 m/s2 , what is the maximum reaction time t of the motorist that will allow him to avoid hitting the deer? Correct answer: 2.28571 s (tolerance 1 %). Explanation: Let : v0 = 14 m/s , v = 0 , and a = -7 m/s2 . x t
Basic Concept: v=
2 v 2 = v0 + 2 a x .
xstop =
2 -v0 = 14 m . 2a x - xstop = v0 t.
homework 04 YOO, HEE Due: Jan 26 2008, 4:00 am This yields t = x - xstop = 2.28571 s . v0 T = 0 m = final separation vt = vT = 26 m/s vc = 41 m/s vC = final speed of car vr = initial relative speed of car to train vR = final relative speed of car to train a = 4 m/s2
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Question 6, chap 2, sect 5. part 2 of 2 10 points If his reaction time is 2.47461 s, how fast will he be traveling when he reaches the deer? Correct answer: 6.08475 m/s (tolerance 1 %). Explanation: Let: t = 2.47461 s Solution: xslow = x - v0 t . We again use the relation
2 v 2 = v0 + 2 a xslow ,
We need to calculate the velocity of the car when it passes the train. In the frame of reference of the train, the initial speed of the car vr relative to the train is vr = vc - vt = 41 m/s - 26 m/s = 15 m/s (1)
When the car reaches the train, the speed of the car vR relative to the train is vR = vC - vt (2)
The velocity vR can be obtained from the kinematic relation
2 2 vR - vr = 2 a
which gives v=
2 v0 + 2 a (x - v0 t)
(3)
= 6.08475 m/s . Question 7, chap 2, sect 5. part 1 of 1 10 points A train is moving parallel and adjacent to a highway with a constant speed of 26 m/s. Initially a car is 55 m behind the train, traveling in the same direction as the train at 41 m/s, and accelerating at 4 m/s2 . What is the speed of the car just as it passes the train? Correct answer: 51.7876 m/s (tolerance 1 %). Explanation: Basic Concepts: Solution: Let (using lower case to be initial and upper case to be final) tt = 0 s = initial time t = tT = final time = t = 55 m
Where a = 4 m/s2 , = 55 m , and vr = 15 m/s from eqn (1). Thus, vR = =
2 vr + 2 a
(4)
(15 m/s)2 + 2(4 m/s2 )(55 m)
= 25.7876 m/s The velocity of the car with respect to the ground just after passing the train is then, from the relation obtained in eqn (2), = vC vR + vt = 25.7876 m/s + 26 m/s = 51.7876 m/s (5)
Working the problem this way gives you experience in making coordinate transformations (changing into the coordinate system of the moving train). This makes the problem easy since it consists of a few simple steps. Many ideas in physics are more easily explained in a different coordinate system from that in which we may be used to.
homework 04 YOO, HEE Due: Jan 26 2008, 4:00 am Alternative explanation: To determine the velocity of the car at the moment when it just passes the train: For simplicity, we can always set t0 = 0 and set the origin of the coordinate system to be at the position of the car. Then the train is at xt = = 55 m . At a later time t, the position of the car will be given by: a (6) xC = vc t + t2 2 while the position of the train is given by: xT = + vt t (7) Basic Concept: aavg = since vf = 0 m/s. Given: vi,1 = +34 m/s vf,1 = 0 m/s a1 = -2.4 m/s2 Solution: t1 = -vi,1 a1 -34 m/s = -2.4 m/s2 = 14.1667 s vf - vi -vi = t t
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When the car catches up with the train at t, one must have xC = xT , or, vc t + a 2 t = + vt t 2 (8)
a 2 t - [vt - vc ] t - = 0 (9) 2 Solve this quadratic equation for the time t vt - vc + (vt - vc )2 + 4 a 2 a
Question 9, chap 2, sect 5. part 2 of 3 10 points Assume that the driver of the chasing car applies the brakes at the same time as the driver of the lead car. b) What must the chasing car's minimum negative acceleration be to avoid hitting the lead car? Correct answer: -3.77177 m/s2 (tolerance 1 %). Explanation: Basic Concepts: The equations simplify to 1 vi x = vf + vi t = t 2 2 and 2 0 = vi + 2ax since vf = 0 m/s. Given: vi,2 vf,2 x2 t2 = +47 m/s = 0 m/s = x1 + 52 m = t1 = 14.1667 s
t=
(10)
= 2.6969 s .
At this moment t, the velocity of the car vC is then given by: vC = vc + a t (11) 2 vC = (41 m/s) + (4 m/s )(2.6969 s) = 51.7876 m/s where vc is the velocity of the car at t = 0. Question 8, chap 2, sect 5. part 1 of 3 10 points Two cars are traveling along a straight line in the same direction, the lead car at 34 m/s and the other car at 47 m/s. At the moment the cars are 52 m apart, the lead driver applies the brakes, causing the car to have an acceleration of -2.4 m/s2 . a) How long does it take for the lead car to stop? Correct answer: 14.1667 s (tolerance 1 %). Explanation:
Solution: The lead car traveled x1 = 34 m/s (14.1667 s) = 240.833 m 2
homework 04 YOO, HEE Due: Jan 26 2008, 4:00 am The chasing car traveled x2 = 240.833 m + 52 m = 292.833 m with a minimum acceleration of a2 =
2 -vi,2
5
7 6 velocity (m/s) 5 4 3 2 1 0 -1 4 5 6 7 8 time (s) What is the position at 9 seconds? Correct answer: 40 m (tolerance 1 %). -2 1 2 3 9
2x2 -(47 m/s)2 = 2(292.833 m) = -3.77177 m/s2
Question 10, chap 2, sect 5. part 3 of 3 10 points c) How long does it take the chasing car to stop? Correct answer: 12.461 s (tolerance 1 %). Explanation: Basic Concepts: a= since vf = 0 m/s. Given: vi,2 = 47 m/s vf - vi -vi = t t
Explanation: The initial position given in the problem is 10 m. The position at 9 seconds is area under the curve from 0 to 9 seconds plus the inital position. It can be calculated by breaking up the time interval from 0 to 9 seconds into smaller time intervals as follows... 7 6 5 4 velocity (m/s) 3 2 1 0 1 2 3 4 5 6 time (s) 7 8 9
a2 = -3.77177 m/s2 Solution: t2 = -vi,2 a2 -47 m/s = -3.77177 m/s2 = 12.461 s
-1 -2
Question 11, chap 2, sect 5. part 1 of 1 10 points Consider the plot below describing motion along a straight line with an initial position of x0 = 10 m.
The position at 2 seconds is 10 meters plus the area of the triangle (shown in gray in the above plot) x = (10 m) + 1 [(2 s) - (0 s)] 2 [(7 m/s) - (0 m/s)] = 17 m ;
homework 04 YOO, HEE Due: Jan 26 2008, 4:00 am however, it can also be calculated: 1 x = xi + vi (tf - ti ) + (tf - ti )2 2 = (10 m) + (0 m/s) [(2 s) - (0 s)] 1 + (3.5 m/s2 ) [(2 s) - (0 s)]2 2 = 17 m . The acceleration during the time interval from 2 to 6 seconds is v a= t (6 m/s) - (7 m/s) = (6 s) - (2 s) = -0.25 m/s2 . The position at 6 seconds is 17 m plus the area of the trapezoid 1 x = (17 m) + [(6 s) - (2 s)] 2 [(6 m/s) + (7 m/s)] = 43 m , however it can also be calculated 1 x = xi + vi (tf - ti ) + (tf - ti )2 2 = (17 m) + (7 m/s) [(6 s) - (2 s)] 1 + (-0.25 m/s2 ) [(6 s) - (2 s)]2 2 = 43 m , where the position is 43 m plus the area of the trapezoid: one-half the base [(6 s) - (2 s)] the sum of the heights [(6 m/s) + (7 m/s)]. The acceleration during the time interval from 6 to 9 seconds is v a= t (-2 m/s) - (0 m/s) = (9 s) - (6 s) = -0.666667 m/s2 . Finally, the position at 9 seconds is 43 m plus the area of the triangle 1 x = (43 m) + [(9 s) - (6 s)] 2 [(-2 m/s) - (0 m/s)] = 40 m , however it can also be calculated x = xi + vi (tf - ti ) + 1 (tf - ti )2 2 = (43 m) + (0 m/s) [(9 s) - (6 s)] 1 + (-0.666667 m/s2 ) [(9 s) - (6 s)]2 2 = 40 m .
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This problem has a different plot for each student.