2. A car at the initial speed of 5 m/s is being uniformly accelerated to a speed of 15 m/s for 5 seconds. Find (a) the acceleration magnitude, (b) the distance the car has travelled during the 5 second period. 3. An object is freely falling from the top of 120 m building, Find (a) the final speed when it reaches the ground, (b) Find the time will it take to fall from the top of the building to the ground. Hollol 4. The apparent weight of a person in an elevator being accelerated upward at a constant acceleration will be (a) smaller than that of his normal weight W= mg (b) equal to his normal weight W = mg, (c) greater than his normal weight W = mg

Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
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Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter2: Motion In One Dimension
Section: Chapter Questions
Problem 2.81CP: A blue car of length 4.52 m is moving north on a roadway (hat intersects another perpendicular...
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Can you solves questions 2-6
1.Linear motion at constant velocity (no acceleration) or at a constant acceleration
(a = fixed value = a constant, V will be changed by a and V is NOT a constant)
For constant velocity V➜ d= V(t)
For constant acceleration a➜ Three (3) equations are being used as follows
Vf= V₁ + a(t); d = Vi (t) + 2 (a) (t²); V² = V₁² + 2(a)(d)
Speed distance / time; velocity magnitude = displacement /time
1. A person walks to the North direction for 120 m in 1 minute, then walk to the East direction
for 160 m in 1 min and 40 seconds. Find (a) the total distance (b) the magnitude of the
displacement, (c) the average speed, and (d) the magnitude of the average velocity.
2. A car at the initial speed of 5 m/s is being uniformly accelerated to a speed of 15 m/s for
5 seconds. Find (a) the acceleration magnitude, (b) the distance the car has travelled during
the 5 second period.
3. An object is freely falling from the top of 120 m building,
Find (a) the final speed when it reaches the ground, (b) Find the time will it take to fall from
the top of the building to the ground.
4. The apparent weight of a person in an elevator being accelerated upward at a constant
acceleration will be (a) smaller than that of his normal weight W= mg
(b) equal to his normal weight W = mg, (c) greater than his normal weight W = mg
II. Two dimensional motion
Projectile motion
Initial horizontal and vertical velocity
Vix = Vi [ cose ];
Viy= Vi [sine ]
Final horizontal and vertical velocity
Vix Vix =Vi [ cose ];
Vfy = Viy-g(t)
0= 45° will have a maximum horizontal displacement (X dirction)
5. A projectile with an initial velocity magnitude of 10 m/s, the initial velocity is making a
30° angle above the horizontal line. Find (a) the initial horizontal component of the initial
velocity (b) the initial vertical component of the initial velocity, (c) how much time will the
projectile take to fly up to the maximum height?
Horizonal Uniform Circular Motion
V=2π (r)/ (t), ac = V²/(r);
Fe = mac
6. An 2 kg object is making a horizontal uniform circular motion. The radius of the circle is 2
meters and it makes 12 revolutions per minute. Find (a) it speed, (b) the centripetal
acceleration magnitude of this object, (c) the centripetal force magnitude on the object.
III. Impulse, momentum, and conservation of momentum
Impulse=J= F (A t);
momentum = P = MV
Impulse - Momentum theorem J=AP impulse causes momentum change
= MV2-MV₁
Conservation of momentum
Before collision, total momentum P= P1 + P2 = M₁ V₁ +M₂V₂
After Collision, total momentum P' = P₁' + P2' = M₁V₁' + M₂V2'
Transcribed Image Text:1.Linear motion at constant velocity (no acceleration) or at a constant acceleration (a = fixed value = a constant, V will be changed by a and V is NOT a constant) For constant velocity V➜ d= V(t) For constant acceleration a➜ Three (3) equations are being used as follows Vf= V₁ + a(t); d = Vi (t) + 2 (a) (t²); V² = V₁² + 2(a)(d) Speed distance / time; velocity magnitude = displacement /time 1. A person walks to the North direction for 120 m in 1 minute, then walk to the East direction for 160 m in 1 min and 40 seconds. Find (a) the total distance (b) the magnitude of the displacement, (c) the average speed, and (d) the magnitude of the average velocity. 2. A car at the initial speed of 5 m/s is being uniformly accelerated to a speed of 15 m/s for 5 seconds. Find (a) the acceleration magnitude, (b) the distance the car has travelled during the 5 second period. 3. An object is freely falling from the top of 120 m building, Find (a) the final speed when it reaches the ground, (b) Find the time will it take to fall from the top of the building to the ground. 4. The apparent weight of a person in an elevator being accelerated upward at a constant acceleration will be (a) smaller than that of his normal weight W= mg (b) equal to his normal weight W = mg, (c) greater than his normal weight W = mg II. Two dimensional motion Projectile motion Initial horizontal and vertical velocity Vix = Vi [ cose ]; Viy= Vi [sine ] Final horizontal and vertical velocity Vix Vix =Vi [ cose ]; Vfy = Viy-g(t) 0= 45° will have a maximum horizontal displacement (X dirction) 5. A projectile with an initial velocity magnitude of 10 m/s, the initial velocity is making a 30° angle above the horizontal line. Find (a) the initial horizontal component of the initial velocity (b) the initial vertical component of the initial velocity, (c) how much time will the projectile take to fly up to the maximum height? Horizonal Uniform Circular Motion V=2π (r)/ (t), ac = V²/(r); Fe = mac 6. An 2 kg object is making a horizontal uniform circular motion. The radius of the circle is 2 meters and it makes 12 revolutions per minute. Find (a) it speed, (b) the centripetal acceleration magnitude of this object, (c) the centripetal force magnitude on the object. III. Impulse, momentum, and conservation of momentum Impulse=J= F (A t); momentum = P = MV Impulse - Momentum theorem J=AP impulse causes momentum change = MV2-MV₁ Conservation of momentum Before collision, total momentum P= P1 + P2 = M₁ V₁ +M₂V₂ After Collision, total momentum P' = P₁' + P2' = M₁V₁' + M₂V2'
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