
A car is moving along a straight road with a speed of 144 km/hr brought to stop within a distance of 200m. How long does it take for the car to stop?
(A) 5s
(B) 10s
(C) 15s
(D) 20s
Answer
586.5k+ views
Hint: Here initially the car is moving and finally, it comes to rest. So negative acceleration called retardation takes place. Stopping distance is also given and we have to find the stopping time.
Complete step by step answer:
The initial velocity is given in km/h and we first convert it into m/s
Initial velocity,
\[\begin{align}
& u=144km/h \\
& =144\times \dfrac{5}{18}m/s \\
& =40m/s \\
\end{align}\]
Final velocity, v=0
Stopping distance, s= 200m
Using the third equation of motion we get
\[\begin{align}
& {{v}^{2}}-{{u}^{2}}=2as \\
& 0-{{40}^{2}}=2\times a\times 200 \\
& -1600=400a \\
& a=-4m{{s}^{-2}} \\
\end{align}\]
So, the acceleration comes out to be, \[a=-4m{{s}^{-2}}\]
Now we can use the first equation of motion,
$
v = u + at \\
\Rightarrow 0 = 40 - 4t \\
\Rightarrow -4t = -40 \\
\Rightarrow 4t = 40 \\
\Rightarrow t = 10 s \\
$
Hence, the time taken by the car after the application of brakes to stop is 10 s and it covers 200m during these 10s.
So, the correct answer is “Option B”.
Note:
While doing problems involving Newton’s equations of motion, we have to keep in mind the sign conventions. If a body is accelerating then its acceleration is positive and if the body is coming to stop after some time then it is accelerating and its acceleration is negative. Also, all the units to be used must be in SI units to avoid any mistake. This has to be particularly noted that time can never be negative and if in our answer time is coming in negative then it means we have committed a mistake.
Complete step by step answer:
The initial velocity is given in km/h and we first convert it into m/s
Initial velocity,
\[\begin{align}
& u=144km/h \\
& =144\times \dfrac{5}{18}m/s \\
& =40m/s \\
\end{align}\]
Final velocity, v=0
Stopping distance, s= 200m
Using the third equation of motion we get
\[\begin{align}
& {{v}^{2}}-{{u}^{2}}=2as \\
& 0-{{40}^{2}}=2\times a\times 200 \\
& -1600=400a \\
& a=-4m{{s}^{-2}} \\
\end{align}\]
So, the acceleration comes out to be, \[a=-4m{{s}^{-2}}\]
Now we can use the first equation of motion,
$
v = u + at \\
\Rightarrow 0 = 40 - 4t \\
\Rightarrow -4t = -40 \\
\Rightarrow 4t = 40 \\
\Rightarrow t = 10 s \\
$
Hence, the time taken by the car after the application of brakes to stop is 10 s and it covers 200m during these 10s.
So, the correct answer is “Option B”.
Note:
While doing problems involving Newton’s equations of motion, we have to keep in mind the sign conventions. If a body is accelerating then its acceleration is positive and if the body is coming to stop after some time then it is accelerating and its acceleration is negative. Also, all the units to be used must be in SI units to avoid any mistake. This has to be particularly noted that time can never be negative and if in our answer time is coming in negative then it means we have committed a mistake.
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