
A car is moving with speed u. Driver of the car sees a red traffic light. His reaction time is t, then find out the distance travelled by the car after the instant when the driver decided to apply brakes. Assume uniform retardation after applying brakes.
Answer
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Hint: In this question we have been asked to calculate the total distance travelled by the car before coming to rest. We have been given that the driver has a reaction time t. We know that the car will have travelled some distance when the diver sees the red light and some distance is covered after applying brakes. Therefore, to calculate the total distance we shall calculate the distance travelled at these two instances. The sum of the two distances shall be our answer.
Formula Used: \[{{v}^{2}}={{u}^{2}}+2as\]
Complete answer:
When the driver sees the red light he takes time to hit the brakes. This is known as the reaction time. Now, we know that the car will have covered some distance, say \[{{S}_{1}}\] during the reaction time because of the initial velocity u. We know that speed is given as distance over time.
Therefore, the distance covered during the reaction time t is,
\[{{S}_{1}}=ut\] ……………. (1)
Now, when the brakes are applied and the car starts slowing down due to retardation a, the car will cover some more distance say \[{{S}_{2}}\]before coming to rest.
Therefore, distance covered during retardation is given by,
\[{{v}^{2}}={{u}^{2}}+2as\]
The final velocity shall be zero as the car is at rest now.
On solving the above equation.
\[-{{u}^{2}}=2a{{S}_{2}}\]
Therefore,
\[{{S}_{2}}=\dfrac{-{{u}^{2}}}{2a}\]
Here the distance is negative due to the negative acceleration. Therefore, we shall consider only the magnitude of the distance.
Therefore,
\[\left| {{S}_{2}} \right|=\dfrac{{{u}^{2}}}{2a}\] ………… (2)
Now the total distance is given by,
\[S={{S}_{1}}+{{S}_{2}}\] ………….. (3)
Now, from (1), (2) and (3)
We get,
\[S=ut+\dfrac{{{u}^{2}}}{2a}\]
Therefore, the total distance travelled will be given by, \[S=ut+\dfrac{{{u}^{2}}}{2a}\]
Note:
The equations of motion are a set of equations that are used to describe the motion of a particle or system. These equations relate the parameters such as initial and final velocity, acceleration, distance and time. The equations of motion are applicable to any system having constant acceleration or deceleration.
Formula Used: \[{{v}^{2}}={{u}^{2}}+2as\]
Complete answer:
When the driver sees the red light he takes time to hit the brakes. This is known as the reaction time. Now, we know that the car will have covered some distance, say \[{{S}_{1}}\] during the reaction time because of the initial velocity u. We know that speed is given as distance over time.
Therefore, the distance covered during the reaction time t is,
\[{{S}_{1}}=ut\] ……………. (1)
Now, when the brakes are applied and the car starts slowing down due to retardation a, the car will cover some more distance say \[{{S}_{2}}\]before coming to rest.
Therefore, distance covered during retardation is given by,
\[{{v}^{2}}={{u}^{2}}+2as\]
The final velocity shall be zero as the car is at rest now.
On solving the above equation.
\[-{{u}^{2}}=2a{{S}_{2}}\]
Therefore,
\[{{S}_{2}}=\dfrac{-{{u}^{2}}}{2a}\]
Here the distance is negative due to the negative acceleration. Therefore, we shall consider only the magnitude of the distance.
Therefore,
\[\left| {{S}_{2}} \right|=\dfrac{{{u}^{2}}}{2a}\] ………… (2)
Now the total distance is given by,
\[S={{S}_{1}}+{{S}_{2}}\] ………….. (3)
Now, from (1), (2) and (3)
We get,
\[S=ut+\dfrac{{{u}^{2}}}{2a}\]
Therefore, the total distance travelled will be given by, \[S=ut+\dfrac{{{u}^{2}}}{2a}\]
Note:
The equations of motion are a set of equations that are used to describe the motion of a particle or system. These equations relate the parameters such as initial and final velocity, acceleration, distance and time. The equations of motion are applicable to any system having constant acceleration or deceleration.
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