Question

A motorboat 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)

Answer 1

Answer:

a. $c  =  0.1739 \frac{m}{s^2}  $

b. $v(40.0 s ) = 1.64 \frac{m}{s}$

Explanation:

So, the equation for the speed is:

$v(t) = v_i - c t$

as we know that the initial speed is

$v_i = 8.2 \frac{m}{s}$

and the speed at t=18.4 s is

$v(18.4 s ) = 8.2 \frac{m}{s}- c 18.4 s = 5.00 \frac{m}{s}$

Now, we can work it a little:

$- c 18.4 s = 5.00 \frac{m}{s} - 8.2 \frac{m}{s}$

$- c 18.4 s = -3.20 \frac{m}{s}$

$- c 18.4 s = -3.20 \frac{m}{s}$

$c  = \frac{ -3.20 \frac{m}{s} }{ - 18.4 s }$

$c  =  0.1739 \frac{m}{s^2}  $

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

$v(40.0 s ) = 8.2 \frac{m}{s}- 0.1739 \frac{m}{s^2} * 40.0 s$

$v(40.0 s ) = 8.2 \frac{m}{s}- 6.96 \frac{m}{s}$

$v(40.0 s ) = 1.64 \frac{m}{s}$