What is the acceleration and the velocity of maximum height? Write the answers i

Question

What is the acceleration and the velocity of maximum height? Write the answers in standard unit VECTOR form. 13. an object thrown straight up, at the instant the object reaches 14. An object accelerates with a uniform acceleration a- 12 m/s. Find the velocity and position at t-20 sec, if the initial velocity is +20 m/s and the initial position is 80 m. Calculate the orbital velocity and the centripetal acceleration of the Space Shuttle if the vehicle orbits at an elevation of 300 km and the orbital period is 90 minutes. 15. Part II. (10 pts each) The equation ofrmotion of a moving object is: x(t)-50-40t+ 2t-0.25t 1· a) Write the equations for velocity, v(t), and acceleration, a(t). b) Find the times at which x, v, and a have their maximum values. c) Find the average velocity between t-2 sec and t-4 sec.

Explanation / Answer

13.

Initial velocity = vi

Final velocity = vf

At maximum height ,

vfx = 0 m/s……….along horizontal direction

vfy = 0 m/s……….along horizontal direction

Thus at maximum height vf = (0i^+0j^)m/s

Object thrown is in the free fall situation, hence acceleration on the object = acceleration due to gravity = a = g ….in upward direction

Thus

a= (0i^ -9.8j^) m/s2

14.

a=-12m/s2, vi=20m/s, x0=80m, t=20s

vf=vi+at

vf= 20-12*20

vf= -220 m/s

xf= x0 + vi*t +1/2*at2

xf= 80+20*20+1/2*-12*20^2

xf= -1920 m

15.

Space shuttle orbits around the earth

T = 90min = 90*60s = 5400s

h=300km

r=radius of earth = 6371km

R=r+h = 6371+300 = 6671km = 6.67*10^6 m

Orbital velocity = distance travelled/ period

v=2R/T

v=(2*3.14*6.67*10^6)/5400

v= 7.76*10^3 m/s

Centripetal acceleration = ac = v2/R

ac = (7.76*10^3)^2/(6.67*10^6) =

ac = 9.03 m/s2

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