Amotorboat cuts its engine when its speed is 8.2 m/s and coasts to rest. the equation describing the motion of the motorboat during this period is v = vie-ct, where v is the speed at time t, vi is the initial speed at t = 0, and c is a constant. at t = 18.4 s, the speed is 5.00 m/s.

(a) find the constant c (in s-1)
(b) what is the speed at t = 40.0 s? (in m/s)

a.

b.

Explanation:

So, the equation for the speed is:

as we know that the initial speed is

and the speed at t=18.4 s is

Now, we can work it a little:

So, at t=40.0 s the speed will be:

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form