University:
Georgia Southern University
Course:
MATH 2242 | Calculus II
Academic year:

2024
Views:

38

Pages:

12
Author:

Jarrett Brock

Question 1 A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s 10.0s of its motion, the vertical acceleration of the rocket is given by 𝑎 𝑦 = 12.80 𝑡 m/s 2 a y =12.80tm/s 2 , where the + +-direction is upward. (a) What is the height of the rocket above the surface of the earth at 𝑡 = 10.0 s t=10.0s? (b) What is the speed of the rocket when it is 325 m 325m above the surface of the earth? Answer 1 (a) Since the acceleration is constant, integrate the first kinematics equation: 𝑣 𝑦 = ∫ 𝑎 𝑦 𝑑 𝑡 = ∫ 12.80 𝑡 𝑑 𝑡 = 6.4 𝑡 2 + 𝑣 0 . v y =∫a y dt=∫12.80tdt=6.4t 2 +v 0 . 𝑦 = ∫ 𝑣 𝑦 𝑑 𝑡 = ∫ ( 6.4 𝑡 2 ) 𝑑 𝑡 = 2.13 𝑡 3 + 𝑦 0 . y=∫v y dt=∫(6.4t 2 )dt=2.13t 3 +y 0 . Substitute 𝑡 = 10.0 s t=10.0s: 𝑦 = 2.13 ( 10.0 ) 3 = 2133.33 m . y=2.13(10.0) 3 =2133.33m. (b) To find 𝑡 t when 𝑦 = 325 m y=325m, solve: 325 = 2.13 𝑡 3 ⟹ 𝑡 3 = 325 2.13 = 152.58 ⟹ 𝑡 = 152.58 3 ≈ 5.85 s . 325=2.13t 3 ⟹t 3 = 2.13 325 =152.58⟹t= 3 152.58 ≈5.85s. Substitute 𝑡 = 5.85 s t=5.85s into 𝑣 𝑦 = 6.4 𝑡 2 v y =6.4t 2 : 𝑣 𝑦 = 6.4 ( 5.85 ) 2 = 202.92 m/s . v y =6.4(5.85) 2 =202.92m/s. Question 2 An antelope moving with constant acceleration covers the distance between two points 70.0 m 70.0m apart in 6.00 s 6.00s. Its speed as it passes the second point is 15.0 m/s 15.0m/s. (a) What are its speed at the first point and (b) its acceleration? Answer 1 (a) Use equations of motion: 𝑣 𝑥 = 𝑣 0 + 𝑎 𝑡 , 𝑥 = 𝑥 0 + 𝑣 0 𝑡 + 1 2 𝑎 𝑡 2 . v x =v 0 +at,x=x 0 +v 0 t+ 2 1 at 2 . Substitute 𝑥 = 70 m , 𝑣 𝑥 = 15 m/s , 𝑡 = 6.00 s x=70m,v x =15m/s,t=6.00s: 𝑎 = 𝑣 𝑥 − 𝑣 0 𝑡 , 𝑥 = 𝑣 0 𝑡 + 1 2 𝑎 𝑡 2 . a= t v x −v 0 ,x=v 0 t+ 2 1 at 2 . Solve for 𝑎 a and 𝑣 0 v 0 : 𝑣 0 = 2 𝑥 𝑡 − 𝑣 𝑥 = 2 ( 70 ) 6 − 15 = 8.33 m/s . v 0 = t 2x −v x = 6 2(70) −15=8.33m/s. 𝑎 = 𝑣 𝑥 − 𝑣 0 𝑡 = 15 − 8.33 6 = 1.11 m/s 2 . a= t v x −v 0 = 6 15−8.33 =1.11m/s 2 . Question 3 A hot-air balloonist, rising vertically at 5.00 m/s 5.00m/s, releases a sandbag at 40.0 m 40.0m above the ground. (a) Find the position and velocity of the sandbag at 𝑡 = 0.25 s t=0.25s and 𝑡 = 1.0 s t=1.0s. (b) How many seconds after its release does the bag strike the ground? (c) What is the velocity just before impact? (d) What is the maximum height above the ground? Answer 1 (a) Position: 𝑦 = 𝑦 0 + 𝑣 0 𝑡 + 1 2 𝑎 𝑡 2 , 𝑣 = 𝑣 0 + 𝑎 𝑡 . y=y 0 +v 0 t+ 2 1 at 2 ,v=v 0 +at. For 𝑡 = 0.25 s t=0.25s: 𝑦 = 40 + 5 ( 0.25 ) − 4.9 ( 0.25 ) 2 = 40.9 m , 𝑣 = 5 − 9.8 ( 0.25 ) = 2.55 m/s . y=40+5(0.25)−4.9(0.25) 2 =40.9m,v=5−9.8(0.25)=2.55m/s. For 𝑡 = 1.0 s t=1.0s: 𝑦 = 40 + 5 ( 1 ) − 4.9 ( 1 ) 2 = 40.1 m , 𝑣 = 5 − 9.8 ( 1 ) = − 4.8 m/s . y=40+5(1)−4.9(1) 2 =40.1m,v=5−9.8(1)=−4.8m/s. (b) Time to ground: 𝑦 = 0 ⟹ 0 = 40 + 5 𝑡 − 4.9 𝑡 2 ⟹ 𝑡 = − 5 ± 5 2 − 4 ( 4.9 ) ( − 40 ) 2 ( 4.9 ) . y=0⟹0=40+5t−4.9t 2 ⟹t= 2(4.9) −5± 5 2 −4(4.9)(−40) . 𝑡 = 3.41 s . t=3.41s. (c) Final velocity: 𝑣 = 5 − 9.8 ( 3.41 ) = − 28.4 m/s . v=5−9.8(3.41)=−28.4m/s. (d) Maximum height: 𝑡 max = − 𝑣 0 𝑎 = 5 9.8 = 0.51 s . t max = a −v 0 = 9.8 5 =0.51s. 𝑦 max = 40 + 5 ( 0.51 ) − 4.9 ( 0.51 ) 2 = 41.3 m . y max =40+5(0.51)−4.9(0.51) 2 =41.3m. Question 4 A 550 N 550N physics student stands on a scale in an elevator supported by a cable. The combined mass of the elevator and student is 850 kg 850kg. (a) Find the acceleration of the elevator when the scale reads 450 N 450N. (b) Find the tension in the cable. Answer 1 (a) Acceleration: 𝑁 = 𝑚 𝑔 + 𝑚 𝑎 , 𝑎 = 𝑁 − 𝑚 𝑔 𝑚 . N=mg+ma,a= m N−mg . 𝑎 = 450 − 550 56.1 = − 1.78 m/s 2 . a= 56.1 450−550 =−1.78m/s 2 . (b) Tension: 𝑇 = 𝑀 𝑔 + 𝑀 𝑎 , 𝑀 = 850 kg . T=Mg+Ma,M=850kg. 𝑇 = 850 ( 9.8 − 1.78 ) = 6817 N . T=850(9.8−1.78)=6817N.