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A driver of a car traveling at $52\ km h^{-1}$ applies the brakes and accelerates uniformly in the opposite direction. The car stops in $5\ s$. Another driver going at $3\ km h^{-1}$ in another car applies his brakes slowly and stops in $10\ s$. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars traveled farther after the brakes were applied?
Given: A driver of a car traveling at $52\ km h^{-1}$ applies the brakes and accelerates uniformly in the opposite direction. The car stops in $5\ s$. Another driver going at $3\ km h^{-1}$ in another car applies his brakes slowly and stops in $10\ s$.
To do: To plot the speed versus time graphs for the two cars on the same paper. To find the car which traveled farther after the brakes were applied.
Solution:
For car$_1$:
Initial speed $u=52\ kmh^{-1}$ [given]
$=52\times\frac{5}{18}\ m/s$ [On converting km/h into m/s]
$=14.44\ m/s$
Final speed $v=0$
Time $t=5\ s$
Here,
At, $t=0$, speed $=14.44\ m/s$, let us say it point $A(0, 14.44)$.
AT, $t=5$, speed $=0$, let us say it point $B(5, 0)$.
For car$_2$:
Initial speed $u=3\ kmh^{-1}$ [given]
$=3\times \frac{5}{18}\ m/s$ [On converting km/h into m/s]
$=0.83\ m/s$
Time $t=10\ s$
Final speed $v=0$
Here,
At, $t=0\ s$, speed $=0.83\ m/s$, let us say it point $C(0, 0.83)$
At, $t=10\ s$, speed $=0$, let us say it point $D(10, 0)$
Speed-Time graph for both cars:
On plotting points $A(0, 14.44)$ and $B(5, 0)$ on the graph paper we obtain graph $AB$ which represents speed-time graph for car$_1$.
On plotting points $C(0, 0.83)$ and $D(10, 0)$ on graph paper we obtain graph $CD$ which represent speed-time graph for car$_2$.
Distance covered by car$_1=$ Area of $\triangle AOB$
$=\frac{1}{2}\times AO\times OB$
$=\frac{1}{2}\times 14.44\times5$
$=7.22\times5$
$=36.10\ m$
Distance covered by car$_2=$ Area of $\triangle COD$
$=\frac{1}{2}\times CO\times OD$
$=\frac{1}{2}\times 0.83\times 10$
$=0.41\times10$
$=4.1\ m$
Therefore, Car$_1$ moved farther after the brakes applied.
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