Motion in a straight-line end | Bài tập môn Vật lý đại cương 1 CTTT | Trường Đại học Bách Khoa Hà Nội

02/07/2021
Chapter 2. Motion Along a Straight Line displacement
the change in position xAB = xB - xA x distance traveled Chapter 2:
the total length of the path traveled distance traveled
between two positions d  x xB AB B Motion along a straight line 𝑉 displacement M = time
Exercises: 3, 7, 9, 13, 15, 19, 21, 31, 33, 37, 39, 43, 45, 57, 51, 53 x displacement A A
Problems: 57(57), 59(59), 65(65), 67(67), 69(69), 73(73), 75(75), 79(79), 81(81), 83(83), 85(85), 89(89), 93(93) t 0 t t A B Dang Duc Vuong distance traveled Speed =  0 time
Email: vuong.dangduc@hust.edu.vn velocity Acceleration = time One Love. One Future. 1 One Love. One Future. 1 2
2.57. Dan gets on Interstate Highway I-80 at Seward, Nebraska, and drives In general
due west in a straight line and at an average velocity of magnitude 88 km/h.
After travelling 76 km, he reaches the Aurora exit. Realizing he has gone
too far; he turns around and drives due east 34 km back to the York exit at
an average velocity of magnitude 72 km/h. For his whole trip from Seward
to the York exit, what are (a) his average speed and (b) the magnitude of his average velocity? (a) his average speed distance traveled average speed  time
Distance traveled for his whole trip (Seward- Aurora-York): d = 76 + 34 = 110 km  km  76 34 avg. speed  82 
The total elapses time (Seward- Aurora-York) t  t  t  
 0.8636  0.4722 1.336h  h  Motion equation: x(t) SA AY 88 72 b, his average velocity displacement average velocity  time
Displacement = x(York)-x(Seward)= 76-34 = 42 km  km  avg. velocity  31
The total elapses time t = 1.336 h    h  One Love. One Future. One Love. One Future. 4 3 4 1 02/07/2021
2.59. A world-class sprinter accelerates to his maximum speed in 4.0 s. He then maintains this speed for the remainder of a
65. A ball starts from rest and rolls down a hill with uniform acceleration, traveling 150 m during the second 5.0 s of its
100m race, finishing with a total time of 9.1 s. (a) What is the runner's average acceleration during the first 4.0 s? (b) What
motion. How far did it roll during the first 5.0 s of motion? y t
is his average acceleration during the last 5.1 s? (c) What is his average acceleration for the entire race? (d) Explain why 1= 5s x1= ? t
your answer to part (c) is not the average of the answers to parts (a) and (b). 2= 5s t x
a, During he accelerates to his maximum speed: 0= 0s x2= 150 m 1 t1= 4.0 s t2= 9.1s v t1= 5s 1 v  v  a t x x 0 x x 1 v  a t 2 1 x 1 2 x  x  v t  a t 1 t 0 0 x x 2= 10 s x 2 0 x 2  1 x a t x x 2=100 m x  0;v  0 1 x 1 0 0 x 2 During he maintains speed (a
uniform acceleration  ax = constant x=0): 1 2x 2100  m  1
x  x  v t  x  v (t  t ) 2
 x  a t  a t (t  t ) 2  a    3.5 
Motion equation of ball along x-axis 𝑥 = 𝑥 + 𝑣 𝑡 + 𝑎 𝑡 2 1 1 1 1 2 1 2 x 1 x 1 2 1 2 x 2 2 2
t  2t (t  t ) 4.0  2 4.0  (9.1 4.0)  s  2 1 1 2 1
b, His average acceleration during the last 5.1 s: a 1 x = 0. 𝑥 = 0; 𝑣 = 0 ⇒ 𝑥 = 𝑎 𝑡 2 V  elocity v a t 3.5 4.0  m  1
c, His average acceleration for the entire race? 1 x 1 a      1.5  avg 2 time t t 9.1  s  𝑥 = 𝑎 𝑡 2.x 2150  m  2 1 2     2 a 4.0
𝛥𝑥 = 𝑎 (𝑡 − 𝑡 )   x 2 2 2 2 2
(d) Explain why your answer to part (c) is not the average of the answers to parts (a) and (b). 1 (t  t ) 10  5  s  𝑥 = 𝑎 𝑡 2 2 1 1 1 1
The runner spends different times moving with the average accelerations of parts (a) and (b). 2
𝛥𝑥 = 𝑎 (𝑡 − 𝑡 ) = 𝑎 𝑡 2
  4.05  50.0m 2 2 2 One Love. One Future. 5 One Love. One Future. 5 6
2.67 Large cockroaches can run as fast as 1.50 m/s in short bursts. Suppose you turn on the light in a cheap motel and see
2.69 An automobile and a truck start from rest at the same instant, with the automobile initially at some distance behind
one scurrying directly away from you at a constant 1.50 m/s. If you start 0.90 m behind the cockroach with an initial speed
the truck. The truck has a constant acceleration of 2.10 m/s2, and the automobile an acceleration of 3.40 m/s2. The
of 0.80 m/s toward it, what minimum constant acceleration would you need to catch up with it when it has traveled 1.20 m,
automobile overtakes the truck after the truck has moved 40.0 m. (a) How much time does it take the automobile to
just short of safety order a counter?
overtake the truck? (b) How far was the automobile behind the truck initially? (c) What is the speed of each when they 0.90 m
are abreast? (d) On a single c, sketch the position of each vehicle as a function of time. Take x = 0 at the initial location
Take x = 0 to be at the t = 0 location of the roach and positive x of the truck.
to be in the direction of motion of the two objects. a1 = ? a2= 2.1 m/s2 a1= 3.4 m/s2 You: V10 = 0.80 m/s Car: 1 1 V 2 2 10=0 V20=0 1 1 a
x (t)  x  v t  a t  x   3.40 t 2 2
x (t)  x  v t  a t  0.80t  a t 2 = 0 1 10 10 1 10 2 2 1 10 10 1 1 2 2 V Truck: 20 = 1.50 m/s x10 X20 =0 x 1 1 Cockroaches 2 2
x (t)  x  v t  a t   2.10t 2 20 20 2 1 2 2 x (t)  x (t) 2
x (t)  x  v t  a t  0.90 1.50t 2 20 20 2
The automobile overtakes the truck after the truck has moved 40.0 m 1 2 2 X x x (t)  x  40 m 2 20  
You catch up with the cockroaches when both objects are at the 10 = 0 20= 0.90 x-axis 2.10 (a) x (t)  x  40 m  t  40  t  6.17(s) 2 20   2
same place at the same time x (t)  x (t) 2 1 2 1 3.40 2 2.10 2 2
0.80t  a t  0.90 1.50t (b) x (t)  x (t)  x  t 
 t  40  x  24.8(m) 1 2 10 10 1 2 2 2 2
you need to catch up with roach when it has traveled 1.20 m: x (t)  x 1.20 m 1.50t  1.20  t  0.8s  a  4.6 m / s
(c) they are abreast at t = 6.17 s  v  v  a t  21.0 m / s ;v  v  a t 13.0 m / s 1 10 1   2 20 2   1   2 20
Minimum constant acceleration: 4.6 m/s2 (d) Graph One Love. One Future. 7 One Love. One Future. 7 8 2 02/07/2021
2.73. Passing. The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.0 m/s (about 45 mile/h). 1 1
Initially, the car is also traveling at 20.0 m/s and its front bumper is 24.0 m behind the truck's rear bumper. The car accelerates The car: 𝑥 = 𝑥 + 𝑣
𝑡 + 𝑎 𝑡 = 0 + 20𝑡 + 0.600𝑡 2 2 v10x= 20 m/s a
at a constant 0.600 m/s2, then pulls back into the truck's lane when the rear of the car is 26.0 m ahead of the front of the truck. 1 2= 0 m/s2 The truck: 𝑥 = 𝑥 + 𝑣
𝑡 + 𝑎 𝑡 = (4.5 + 24.0) + 20𝑡 a1= 600 m/s2 v20x= 20 m/s
The car is 4.5 m long and the truck is 21.0 m long. (a) How much time is required for the car to pass the truck? (b) What 2 * *
distance does the car travel during this time? (c) What is the final speed of the car?
𝑥 − 𝑥 = 26.0 + 21.0 = 47.0m 0 x 4.5 24.0 21.0
Solve equations: (a) t* = 15.9 s v10x= 20 m/s a2= 0 m/s2
(b) distance the car travel during t* =15.9 s: 𝑥 − 𝑥
= 20 t ∗ + 0.600𝑡 ∗ = 394 m a1= 600 m/s2 v v 20x= 20 m/s 20x= 20 m/s (c) the final speed of the car * * 𝑣 (𝑡 ∗) = 𝑣
+ 𝑎 𝑡 ∗= 20 + 0.600 × 15.9 = 29.5 𝑚 𝑠 ⁄ 0 x 0 * * x 4.5 24.0 21.0 4.5 21.0 26.0
Take x = 0 at the initial location of the car before the car accelerates; position of car and truck note by (*) symbol; origin of time t0 = 0. 1 1 The car: 𝑥 = 𝑥 + 𝑣
𝑡 + 𝑎 𝑡 = 0 + 20𝑡 + 0.600𝑡 2 2
At time t, the car to pass the truck 1
𝑥 − 𝑥 = 26.0 + 21.0 = 47.0m The truck: 𝑥 = 𝑥 + 𝑣
𝑡 + 𝑎 𝑡 = (4.5 + 24.0) + 20𝑡 2 One Love. One Future. One Love. One Future. 10 9 10
2.75 The acceleration of a particle is given by a
2.79 Visitors at an amusement park watch divers step off a platform 21.3 m (70 ft) above a pool of water. According to the x(t) = -2.0m/s2 + (3.00m/s3)t.
announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). Air resistance may be ignored. (a) Is the announcer
(a) Find the initial velocity v0x, such that the particle will have the same x-coordinate at t = 4.00 s as it had at t = 0.
correct in this claim? (b) Is it possible for a diver to leap directly upward off the board so that, missing the board on the way
(b) What will be the velocity at t = 4.00 s?
down, she enters the water at 25.0 m/s? If so, what initial upward speed is required? Is the required initial speed physically y-axis
(a) Find the initial velocity v attainable? y-axis 0x g t
(a) The speed of the divers when he enters water: vt  v  a  t 2 dt  v  2.0t 1.5t 0 0 1 Motion equation of divers: 2 y(t)  y  v t  a t 0 0 0 y 2 y vi t with y  21.3 ; m v  0;a  9  .8 m s y y  2 / 0 0  xt  x  v  t 2 3
dt  x  v t  t  0.5t 0 0 0 9.81 The divers enter the water: 2 y(t)  0  21.3  t  0  t  2.08 s f f   0 2 x  x  v  4.0 m / s
v (t)  v  a t  v  9
 .81 2.08  20.4 m / s y 0y y f   t0 t4 0  
(b) What will be the velocity at t = 4.00 s?
The announcer has exaggerated the speed of the diver.
t  4.0s  vt  4.0s  12.0m / s
(b) what initial upward speed vi =? 0 0 1 2 y(t)  y  v t  a t 0 0 y 2 y v  14.4 m / s v 0 y   f vf v (t)  v  a t  2  5.0 y 0y y y  21.3 ; m v  ?;a  9  .8 m s y y  2 / 0 0  One Love. One Future. 11 One Love. One Future. 12 11 12 3 02/07/2021
2.81 A football is kicked vertically upward from the ground and a student gazing out of the window sees it moving upward
2.83 Look Out Below. Sam heaves a l6-lb shot straight upward, giving it a constant upward acceleration from rest of
pass her at 5 m/s. The window is 12m above the ground. Ignore the air resistance.
45.0 m/s2 for 64.0 cm. He releases it 2.20 m above the ground. You may ignore air resistance. (a) What is the speed of the
(a) How high is the ball go above ground?
shot when Sam releases it? (b) How high above the ground does it go? (c) How much time does he have to get out of its
(b) How much time does it take to go from the ground to its highest point?
way before it returns to the height of the top of his head, 1.83 m above the ground? y v y-axis Motion of the shot: 2 Periods 2 Motion equation of ball g g (1) Straight upward: from y 1
1= 64.0 cm to y2 = 2.20 cm; a1= 45.0 m/s2 2 y(t)  y  v t  a t (2) Freely falling body: y 0 0y y 2
20 = 2.20 cm; v20= v2; a2= -9.81 m/s2. v (t)  v  a t (a) What is the speed y 0y y At position of the student v 2 2 release
Period 1: v  v  2a(y  y )  v  7.59 m / s 20 10 20 10 20    y 9.81 v 1 = 5 m/s 20 v 1 20 y(t)  y  12 2 0  v t  t  12 y1 =12 m 1 oy  2  v  16.1m / s v (t)  v  5 0y g = 9.81 m/s2
(b) How high above the ground does it go y 1 v  9.81 t  5  0y 2.20 m v10=0
(a) How high is the ball go above ground? v0 12.0
Period 2: v (t)  v  a t  0  t  0.77s y 2 20 2 10 2 2 v  v  2a (y  y ) 1 2 y 0 y y max 0  m       y  13.3m y (t) y v t a t 5.14m 0.64 m 2 20 20y 2 max v  0 2 a y 
(c) it returns to the height of the top of his head, 1.83 m above the ground y=0
(b) How much time does it take to go from the ground to its highest point? y (t)  1.83  t  1.60s 2
v (t)  v  a t  0  t  1.64s y 0y y One Love. One Future. 13 One Love. One Future. 14 13 14
89. Falling Can. A painter is standing on scaffolding that is raised at constant speed. As he travels upward, he accidentally
2.85 Juggling Act. A juggler performs in a room whose ceiling is 3.0 m above the level of his hands. He throws a ball
nudges a paint can off the scaffolding, and it falls 15.0 m to the ground. You are watching and measure with your
upward so that it just reaches the ceiling. (a) What is the initial velocity of the ball? (b) What is the time required for the ball
stopwatch that it takes 3.25 s for the can to reach the ground. Ignore air resistance. (a) What is the speed of the can just
to reach the ceiling? At the instant when the first ball is at the ceiling. the juggler throws a second ball upward with two-
before it hits the ground? (b) Another painter is standing on a ledge, with his hands 4.00 m above the can when it falls off.
thirds the initial velocity of the first: (c) How long after the second ball is thrown did the two balls pass each other? (d) At
He has lightning-fast reflexes and if the can passes in front of him, he can catch it. Does he get the chance?
what distance above the juggler's hand do they pass each other? + y Motion equation of can: 1 𝑔⃗ Motion equation of ball 1 2 y (t)  y  v t  a t 1 10 10 1 y-axis g 1 2
𝑦 = 𝑦 + 𝑣 𝑡 + 𝑎 𝑡 v (t)  v  a t 2 𝑔 1 10 1 ⇒ 𝑦 = 15 + 𝑣 𝑡 − 𝑡 y  0;a  9  .8  2 1 m / s 10 1  v 𝑣 = 𝑣 2
(a,b) Ball just reaches the ceiling: y v
𝑎 = −𝑔 = −9.8 𝑚 𝑠 ⁄ 1max = 3.0 m; v1max(t)=0. 2 2
The can reach the ground: y = 0 when t = t* = 3.25s  v v  v  2(y  y )  v  7.67 m / s 1max 10 1max 10 10   3 m h 0 = 11.31 m/s 0= 4 m t
(a) The speed of the can just before it hits the ground when t = t* = 3.25s
v (t)  v  a t*  0  t*  0.78s 1 10 1 0= 0 v0 y = 0
v ∗= v + a. t ∗= 11.31 − 9.81 × 3.25 = −20.57 m s ⁄ (c) two balls pass each other (b) Does he get the chance? 1 2 y (t)  y  v t  a t y 1 10 10 1 2 0= 15 m Position of second painter: y 1 1= y0 + h0 = 15 + 4 = 19 m
y (t)  y  v t  t *  a t  t *2 2 20 20 2 with: y 2 20 = 0 m; v20 = 2/3 v10.
Method 1: solve y = y1; 15 + 11.31𝑡 − 𝑡 = 19; have chance 0 t 3.25 s
y (t)  y (t)  t  1.37s  t  t*  0.59s 1 2 (d) y (1.37s)  1.31m 0 t*= 3.25 s
Method 2: find highest position of can (v = 0) ; have chance when ymax  y1 1 One Love. One Future. One Love. One Future. 16 15 16 4 02/07/2021
2.93 Two cars, A and B, travel in a straight line. The distance of A from the starting point is given as a function of time
Two cars, A and B, travel in a straight line. The distance of A from the starting point is given as a function of time by xA(t) = by x
αt + βt2, with α = 2.60 m/s and β = 1.20 m/s2. The distance of B from the starting point is x
A(t) = t + t2, with a = 2.60 m/s and b = 1.20 m/s2. The distance of B from the starting point is xB(t) = t2 - t3, with
B(t) =γt2 - δt3, with γ = 2.80m/s2
 = 2.80m/s2 and  = 0.20m/s3. (a) Which car is ahead just after they leave the starting point? (b) At what time(s) are the
and δ = 0.20m/s3. (a) Which car is ahead just after they leave the starting point? (b) At what time(s) are the cars at the same
cars at the same point? (c) At what time(s) is the distance from A to B neither increasing nor decreasing? (d) At what
point? (c) At what time(s) is the distance from A to B neither increasing nor decreasing? (d) At what time(s) do A and B
time(s) do A and B have the same acceleration? have the same acceleration? 3 2 (c)
AB  0.2t 1.6t  2.6t  const Analyze d  AB 2
(a) Which car leave the starting point first – at t = 0, which car has v > 0? 
 0.6t  3.2t  2.6  0 dt
(b) Two cars at the same position  solving equation x  t 1.00s;t  4.33s 1 2 A(t) = xB(t) with t >0
(a) Which car is ahead just after they leave the starting point
(c) AB=const  d(AB)/dt = 0 or vA = vB (t>0) ; (d) (d) a
(b) At what time(s) are the cars at the same point?
xA= axB d2xA(t)/dt2 = d2xB(t)/dt2 (t>0);
Cars at the same point implies x a  a A B A = xB
 21.2  2 2.8 1.2t  t  2.67s One Love. One Future. 17 One Love. One Future. 18 17 18 www.hust.edu.vn One Love. One Future.
Thank you for your attentions!19 One Love. One Future. 20 19 20 5