The brakes of a car moving at 14m/s are applied, and the car comes to a stop in 4s. (a) what was the cars acceleration? (b) how long would the car take to come to a stop starting from 20m/s with the same acceleration? (c) how long would the car take to slow down from 20m/s to 10m/s with the same acceleration?

(a)

The car's acceleration is given by

where

v = 0 is the final velocity

u = 14 m/s is the initial velocity

t = 4 s is the time elapsed

Substituting,

where the negative sign means the car is slowing down.

(b) 5.7 s

We can use again the same equation

where in this case we have

is again the acceleration of the car

v = 0 is the final velocity

u = 20 m/s is the initial velocity

Re-arranging the equation and solving for t, we find the time the car takes to come to a stop:

(c)

As before, we can use the equation

Here we have

is again the acceleration of the car

v = 10 is the final velocity

u = 20 m/s is the initial velocity

Re-arranging the equation and solving for t, we find

a long strand of spaghetti is folded once. if this is cut 20 times, how many pieces of spaghetti will there be? a long strand of spaghetti is folded once. if this is cut 20 times, how many pieces of spaghetti will there be? a long strand of spaghetti is folded once. if this is cut 20 times, how many pieces of spaghetti will there be? a long strand of spaghetti is folded once. if this is cut 20 times, how many pieces of spaghetti will there be? a long strand of spaghetti is folded once. if this is cut 20 times, how many pieces of spaghetti will there be?

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in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.