PHYS1100 Practice problem set, Chapter 2: 6, 10, 13, 17, 22, 24, 25, 34, 42, 50, 55, 65, 71, 82

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1 PHYS00 Practice problem set, Chapter : 6, 0, 3, 7,, 4, 5, 34, 4, 50, 55, 65, 7, 8.6. Model: We will consider Larry to be a particle. Solve: Since Larry s speed is constant, we can use the following equation to calculate the velocities: sf si vs = t t (a) For the interval from the house to the lamppost: 00 yd 600 yd v = = 00 yd/min 9:07 9:05 For the interval from the lamppost to the tree: 00 yd 00 yd v = = yd/min 9:0 9:07 (b) For the average velocity for the entire run: v avg f 00 yd 600 yd = = + 0 yd/min 9:0 9:05 i

2 .0. Please refer to Figure Ex.0. Solve: (a) We can obtain the values for the velocity-versus-time graph from the equation v = s/ t. (b) There is only one turning point. At t = s the velocity changes from +0 m/s to 0 m/s, thus reversing the direction of motion. At t =, there is an abrupt change in motion from rest to +0 m/s, but there is no reversal in motion..3. Please refer to Figure Ex.3. Solve: (a) (b) The acceleration of the train at t = 3.0 s is + m/s.

3 .7. Model: We are using the particle model for the skater and the kinematics model of motion under constant acceleration. Solve: Since we don t know the time of acceleration we will use Assess: f i v = v + a( x x ) f i f i vf vi (6.0 m/s) (8.0 m/s) a = = =.8 m/s ( x x ) (5.0 m) A deceleration of.8 m/s is reasonable... Model: We will use the particle model and the constant-acceleration kinematic equations. Solve: (a) Substituting the known values into y = y + v t + a t, we get m = 0 m + 0 (m/s) t + ( 9.8 m/s ) t One of the roots of this equation is negative and is not relevant physically. The other root is t = 4.53 s which is the answer to part (b). Using v = v0 + a t, we obtain v = 0(m/s) + ( 9.8 m/s )(4.53 s) = 4.4 m/s (b) The time is 4.53 s. Assess: A time of 4.5 s is a reasonable value. The rock s velocity as it hits the bottom of the hole has a negative sign because of its downward direction. The magnitude of 4.4 m/s compared to 0 m/s, when the rock was tossed up, is consistent with the fact that the rock travels an additional distance of 0 m into the hole. 3

4 .4. Model: We will represent the skier as a particle. Note that the skier s motion on the horizontal, frictionless snow is not of any interest to us. Also note that the acceleration parallel to the incline is equal to g sin0. Solve: Using the following constant-acceleration kinematic equations, v fx = v ix + a x (x f x i ) (5 m /s) = (3.0 m / s) + (9.8 m /s )sin 0 (x 0 m) x = 63.5 m v fx = v ix + a x (t f t i ) (5 m / s) = (3.0 m / s) + (9.8 m /s )(sin 0 )t t = 7.05 s Assess: A time of 7.05 s to cover 63.5 m is a reasonable value Solve: (a) The position t = s is x s = [() + ] m = 5 m. (b) The velocity is the derivative v = dx / dt and the velocity at t = s is calculated as follows: v = (6t ) m/ s v s = [6() ] m /s = 3 m /s (c) The acceleration is the derivative a = dv /dt and the acceleration at t = s is calculated as follows: a = (t) m /s a s = [()()] m/ s = 4 m /s.34. Please refer to Figure P.34. Solve: (a) The velocity-versus-time graph is the derivative with respect to time of the distance-versus-time graph. The velocity is zero when the slope of the position-versus-time graph is zero, the velocity is most positive when the slope is most positive, and the velocity is most negative when the slope is most negative. The slope is zero at t = 0, s, s, 3 s,...; the slope is most positive at t = 0.5 s,.5 s,...; and the slope is most negative at t =.5 s, 3.5 s,... (b) 4

5 .4. Model: Represent the ball as a particle. Please refer to Figure P.4. The ball rolls down the first short track, then up the second short track, and then down the long track. s is the distance along the track measured from the left end (where s = 0). Label t = 0 at the beginning, that is, when the ball starts to roll down the first short track. Solve: Because the incline angle is the same, the magnitude of the acceleration is the same on all of the tracks. Assess: Note that the derivative of the s versus t graph yields the v s versus t graph. And the derivative of the v s versus t graph gives rise to the a s versus t graph. 5

6 .50. Model: The car is a particle and constant-acceleration kinematic equations hold. Solve: (a) This is a two-part problem. During the reaction time, x = x 0 + v 0 (t t 0 ) + / a 0 (t t 0 ) = 0 m + (0 m/s)(0.50 s 0 s) + 0 m = 0 m After reacting, x x = 0 m 0 m = 00 m, that is, you are 00 m away from the intersection. (b) To stop successfully, v = v + a ( x x ) (0 m/s) = (0 m/s) + a (00 m) a = m/s (c) The time it takes to stop can be obtained as follows: v = v + a ( t t ) 0 m/s = 0 m/s + ( m/s )( t 0.50 s) t = 0.5 s 6

.55. Model: We will model the lead ball as a particle and use the constant-acceleration kinematic equations. Note that the particle undergoes free fall until it hits the water surface.

7 .55. Model: We will model the lead ball as a particle and use the constant-acceleration kinematic equations. Note that the particle undergoes free fall until it hits the water surface. Solve: The kinematics equation y = y + v ( t t ) + a ( t t ) becomes Now, once again, 5.0 m = 0 m + 0 m + ( 9.8 m/s )( t 0) t =.0 s y = y + v( t t) + a( t t) y y = v (3.0 s.0 s) + 0 m/s =.99 v v is easy to determine since the time t has been found. Using v = v0 + a0( t t0), we get With this value for v, we go back to: v = 0 m/s (9.8 m/s )(.0 s 0 s) = m/s y y =.99 v = (.99)( m/s) = 9.7 m y y is the displacement of the lead ball in the lake and thus corresponds to the depth of the lake. The negative sign shows the direction of the displacement vector. Assess: A depth of about 60 ft for a lake is not unusual. 7

8 .65. Model: We will use the particle model for the skier s motion ignoring air resistance. Solve: (a) As discussed in the text, acceleration along a frictionless incline is a = gsin5, where g is the acceleration due to gravity. The acceleration of the skier on snow therefore is a ( ) x = 00 m/sin5. The final velocity can now be determined using kinematics = 0.90 gsin5. Also since h/ x = sin5, 00 m v = v0 + a( x x0) = (0 m/s) + (0.90)(9.8 m/s )sin5 0 m sin5 km 3600 s v = m/s = ( m/s) 4 km/ hr 000 m = hr (b) The speed lost to air resistance is (4 80)/4 00% 6%. = Assess: A record of 80 km/hr on such a slope in the presence of air resistance makes the obtained speed of 4 km/hr (without air resistance) physical and reasonable. 8

9 .7. Model: We will represent the dog and the cat in the particle model. Solve: We will first calculate the time t C the cat takes to reach the window. The dog has exactly the same time to reach the cat (or the window). Let us therefore first calculate t C as follows: xc = xc0 + vc0( tc tc0 ) + ac( tc tc0 ) 3.0 m =.5 m + 0 m + (0.85 m/s ) tc tc =.879 s In the time t D =.879 s, the dog s position can be found as follows: xd = xd0 + vd0 ( td td0 ) + ad( td td0) = + + = 0 m (.50 m/s)(.879 s) ( 0.0 m/s )(.879 s).6 m That is, the dog is shy of reaching the cat by 0.4 m. The cat is safe. 9

10 .8. Model: We will use the particle-model to represent the sprinter and the equations of kinematics. Solve: Substituting into the constant-acceleration kinematic equations, x = x + v ( t t ) + a ( t t ) = 0 m + 0 m + a (4 s 0 s) = a t = a (4.0 s) x = (8 s ) a From these two results, we find that x = ( s)v. Now, 0 v = v + a ( t t ) = 0 m/s + a (4.0 s 0 s) v = (4.0 s) a x = x + v( t t) + a( t t) 00 m = ( s) v + v (0 s 4 s) + 0 m v =.5 m/s Assess: Using the conversion.4 mph = m/s, v =.5 m/s = 8 mph. This speed as the sprinter reaches the finish line is physically reasonable. 0

2.1. Model: The car is represented by the particle model as a dot.

2.1. Model: The car is represented by the particle model as a dot. .. Model: The car is represented by the particle model as a dot. Solve: (a) Time t (s) Position x (m) 0 00 975 85 3 750 4 700 5 650 6 600 7 500 8 300 9 0 (b) .. Solve: Diagram Position Velocity Acceleration