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Q. 1 Let the velocity of the dancer as he/she leaves the floor be v. v/0.25 = 10 as only 0.25 s will be available to make the velocity 0. or v = 2.5 m/s
The distance traveled in this time = 2.5*0.25 - 0.5*10*(0.25^2)= 0.625 - 0.3125 = 0.3125 m or 31.25 cm which is much less than required 80 cm hence this is just not possible.
Q. 2. Speed of Rolls Royce = 60000/3600 = 16.67 m/s
Stopping deceleration = [(16.67^2)/(2*240)] = 0.5787 m/s^2
Also we have
deceleration of the truck = [(16.67^2)/(2*50)] = 2.778; time required to decelrate = 16.67 /2.778 = 6.002 or 6 s
Distance traveled by the truck till it stops = 60.5*16.67 + 16.67*6 - 0.5*2.778*36 = 1225.25 m
Distance between stopped truck and Rolls Royce = 1225.25 + 240 = 1465.25 m or 1.465 km
The distance traveled in this time = 2.5*0.25 - 0.5*10*(0.25^2)= 0.625 - 0.3125 = 0.3125 m or 31.25 cm which is much less than required 80 cm hence this is just not possible.
Q. 2. Speed of Rolls Royce = 60000/3600 = 16.67 m/s
Stopping deceleration = [(16.67^2)/(2*240)] = 0.5787 m/s^2
Also we have
deceleration of the truck = [(16.67^2)/(2*50)] = 2.778; time required to decelrate = 16.67 /2.778 = 6.002 or 6 s
Distance traveled by the truck till it stops = 60.5*16.67 + 16.67*6 - 0.5*2.778*36 = 1225.25 m
Distance between stopped truck and Rolls Royce = 1225.25 + 240 = 1465.25 m or 1.465 km
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We will assume that the dancer jumps straigth up in the air. Once the dancer leaves the dance floor, he/she is under the acceleration downward caused by the force of gravity, 10 m/ sec^2. It is helpful to know that it takes an equal time to go up and down. This become a straight line kinemitics problem.
Start at the max height of the dancers (80 cm). initial velocity is 0.
Use dist = 1/2 x a x t^2
0.8 m = 1/2 10 x t^2
solve for t
This must be 1/2 of 0.5 seconds.
Start at the max height of the dancers (80 cm). initial velocity is 0.
Use dist = 1/2 x a x t^2
0.8 m = 1/2 10 x t^2
solve for t
This must be 1/2 of 0.5 seconds.