1. A travels at 25.0 m/s for 2.60 min, motorist drives south 20.0 m/s min, then

Question

1. A travels at 25.0 m/s for 2.60 min, motorist drives south 20.0 m/s min, then turns west and values and finally northwest for 3.00 min trip, find the following travels at 30.0 m/s for 1.00 min. For this 6.60 Total vector displacement, b) average speed, c) average velocity 2 located at the origin has an eleration of 2.00j m/s and an initial velocity of Vi 9.001 m/s. vector position of the particle at any time t (where t is measured in seconds). Find the (a) (b) Find the velocity of the particle at any time t (c) Find the coordinates of the particle at t 9.00 s (d) Find the speed of the particle at t-9.00 s 3. Mayan kings and many school sports teams are named for the puma, cougar, or mountain lion Felis at concolor, the best jumper among animals. It can jump to a height of 13.1 ft when leaving the ground an angle of 41.2°. With what speed, in does it leave the ground to make this leap 4. A cannon with a muzzle m/s is used to start on a mountain slope. The target is 1900 m from the cannon horizontally m above an avalanche the and 795 the cannon At what angle, above horizontal, should the cannon be fired? (gnore air resistance.) 5. We found the centripetal of the Earth as revolves around the Sun. Compute the centripetal acceleration of point on the surface of the Earth at the equator caused by the rotation of the Earth about radius Earth km.) 6. A train slows down as it rounds a sharp horizontal turn, going from 86.0 km/h to 46.0 km/h in the 17.0 s that it takes to round the bend. The radius of the curve is 130 m. Compute the acceleration at the moment the train speed reaches 4 km/h. Assume the train continues to slow down at this time at the same rate. 7. This figure (a -24.5 m/s represents the total acceleration of a particle moving clockwise in a circle of

Explanation / Answer

1. we take SOUTH s +ve x and WEST as +y

for 3 mt 20m/s

x = 20*3*60 = 3600 m

y = 0

next 2.6 m 25 m/s WETS bound

x -dsiplacement =0

y = 25*2.6*60 = 3900 m

next 1 mt - 30 m/s N-W

total displacement = 30*60 = 1800 m

x= -Nort = -1273 m

y = +1273 m

total x dispalcemnt = 3600 -1273 = 2327 m

total y -displacement = 3900 +1273 = 5173 m

displacement vector   r^ = 2327 x^ +5173 y^

average speed = (20*3 +25*2.6 +30*1)/6.6 = 23.48 m/s

avearge velcoity = total displacement/time   = sqrt(23272 + 51732 )/6.6*60   = 18.56 m/s

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