Question

1. a polar bear dives into the water and begins swimming with a speed of 2.60 m/s. by the time it arrives at the shore, its speed has decreased to 2.20 m/s. if the polar bear’s swim takes exactly 9.00 minutes, solve for the following: a) the bear’s acceleration b) the distance traveled by the polar bear

Explanation:

Step1: Convert time to seconds

$t = 9.00\ minutes=9\times60\ s = 540\ s$

Step2: Calculate acceleration

The formula for acceleration is $a=\frac{v - u}{t}$, where $u = 2.60\ m/s$, $v=2.20\ m/s$ and $t = 540\ s$.
$a=\frac{2.20 - 2.60}{540}=\frac{- 0.40}{540}\approx - 7.41\times10^{-4}\ m/s^{2}$

Step3: Calculate distance using the formula $s=ut+\frac{1}{2}at^{2}$

Substitute $u = 2.60\ m/s$, $a=-7.41\times 10^{-4}\ m/s^{2}$ and $t = 540\ s$ into the formula.
$s=(2.60\times540)+\frac{1}{2}\times(-7.41\times 10^{-4})\times(540)^{2}$
$s = 1404- \frac{1}{2}\times7.41\times10^{-4}\times291600$
$s = 1404-108.0$
$s = 1296\ m$

Answer:

a) $-7.41\times 10^{-4}\ m/s^{2}$
b) $1296\ m$