Answer
(a) Car's speed just after leaving the icy portion of the road $=30.4m/s=68 mi/h$
(b) Total distance travels before the car stop $=166 m$
(c) Total time taken by the car to stop $=7.4s$
Work Step by Step
A car moving with a uniform speed of 75 mi/h on a straight road suddenly applies break and comes to rest after travelling some distance. The schematic of the problem is shown in the figure. Before proceeding further into the problem let us convert the speed given in mi/h to m/s. So that all the quantities uses the same system of units.
$1 mi/h=0.447 m/s$
therefore, $ 75 mi/h=75\times 0.447m/s=33.5 m/s$
(a) The car moves with an initial velocity ($u$) of $33.5 m/s$ before entering the icy portion. On the icy portion of the road, the car travels a distance ($y$) of $100m$ and experiences a deceleration($a$) of $1m/s$. let $v$ be the final velocity with which the car leaves the icy portion.
Thus we have,
$u=33.5m/s$
$y=100m$
$a=-1m/s$, since car experience deceleration, the sign of acceleration is negative.
$v= ?$
Using the kinematic equation of motion $v^{2}=u^{2}+2ay$
$v$ can be calculated as $v=\sqrt (u^{2}+2ay)$
substituting the known values, we get
$ v=\sqrt (33.5^{2}+2\times (-1)\times 100)$
$ v=\sqrt (1122.25-200)=\sqrt 922.25=30.4m/s$
Thus, the car leaves the icy portion with a velocity of $30.4 m/s$ or converting back to mi/h gives $(30.4/0.447)=68 mi/h$.
(b) Total distance the car travelled= distance car travelled on the icy road + distance car travelled on the concrete road.
For this, we need to find the distance car travelled on the concrete road. Since the car leaves the icy road with a velocity of $30.4 m/s$, with the same velocity it enters the concrete road.
Therefore, we have initial velocity on the concrete road $u=30.4m/s$.
Deceleration experienced by the car on the concrete road, $a=-7m/s^{2}$.
Let car stop after travelling some distance $y$ on the concrete road. Thus, the final velocity $v=0$.
Using the kinematic equation of motion $v^{2}=u^{2}+2ay$
The distance car travelled on the concrete road $y$ can be calculated as $y=\frac{v^{2}-u^{2}}{2a}$
substituting the known values, we get
$y=\frac{0^{2}-30.4^{2}}{2\times (-7)}=\frac{-30.4^{2}}{-14}=66 m$
Thus, car travels $66m$ on the concrete road before coming to stop.
Total distance travelled by the car $=100m+66m=166m$.
(c) Total time car took = time it took to cover the icy road+time it took on the concrete road before stopping.
Time taken by the car on each section of the road can be calculated using the equation $v=u+at$.
or $t=\frac{v-u}{a}$
For the icy road, $t=\frac{30.4-33.5}{-1}=3.1s$
For the concrete road, $t=\frac{0-30.4}{-7}=4.3s$
Thus, total time taken $=3.1s+4.3s=7.4s$
