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A driver is driving at 50 m/s. Suppose the coefficient of static friction betwee

ID: 1554980 • Letter: A

Question

A driver is driving at 50 m/s. Suppose the coefficient of static friction between the tires and the road is 0.75. (a) If the driver brakes, what is the distance before the car comes to a stop. (b) If the road is wet so that the coefficient of friction goes down to 0.45 but the initial speed is still 50 m/s, what is the distance before the car comes to a stop after application of the brakes. Pluto is 4 billion miles from the sun. Light from the sun travels at 186000 miles per second. How many hours does it take for light to reach from the sun to Pluto? A gymnast springs vertically upward from a trampoline. The gymnast reaches a maximum height of 3.6 metres before falling back down. All heights are measured with respect to the ground. Ignoring air resistance, find the initial speed with which the gymnast leaves the trampoline. A motorcycle has a constant acceleration of 2.5 m/s2. Both the velocity and acceleration of the motorcycle point in the same direction. How much time is required for the motorcycle to change its speed from a) 21 m/s to 31 m/s b) 51 m/s to 61 m/s.

Explanation / Answer

1)

acceleration a= -u*g


vf^2 - vi^2 = 2*a*x

0^2 - 50^2 = -2*0.75*9.8*x

x = 170 m <<<------answer


(b)


acceleration a= -u*g


vf^2 - vi^2 = 2*a*x

0^2 - 50^2 = -2*0.45*9.8*x

x = 283.44 m <<<------answer

====================


(2)


time = distance /speed

distance = 4 billion miles = 4*10^9 miles

time = (4*10^9)/(18600)

time = 215053.76 s


1 hour = 2=60 min = 60*60 s


1 s = 1h/(3600)

time = 215053.76*(1/3600)h = 59.74 hours <<<<<<<-------------answer

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(3)

along vertical


initial velocity = vi

final velocity at maximum height vf = 0

acceleration a = -g = -9.8 m/s^2

displacement y = 3.6 m

vf^2 - vi^2 = 2*a*y

0 - vi^2 = -2*9.8*3.6


vi = 8.4 m/s <<<<-----------answer


================

(4)

acceleration a = (vf-vi)/t


(a)


time = (vf-vi)/a

t = (31-21)/2.5 = 4 s

(b)

t = (61-51)/2.5 = 4 s

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