Two cars A and B start from the same starting point. A is starting from rest and moving with constant acceleration $2m/{s^2}$ and car B is starting with a constant velocity $40m/s$. After what time both cars will meet each other?
Answer
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Hint: The second equation of motion also known as the position time relation is used to calculate the distance travelled by the body. To solve the given question, we will apply the formula of the second equation of motion to find the required answer.
Formula Used: $S = ut + \dfrac{1}{2}a{t^2}$
Complete answer:
The second equation of motion also known as the position time relation is mathematically given as, $S = ut + \dfrac{1}{2}a{t^2}$
Where ‘s’ is the displacement, ‘u’ is the initial velocity, ‘t’ is the time and ‘a’ is the acceleration of the body.
This formula is used to calculate the displacement when the values for time, acceleration and initial velocity are known.
Now,
Let the distance covered by car A be ${S_1}$
Let the distance covered by the car B be ${S_2}$
Both cars will meet each other when ${S_1} = {S_2}$
The position time relation of car A can be given as,
${S_1} = ut + \dfrac{1}{2}a{t^2}$
The position time relation of car B can be given as,
${S_2} = ut + \dfrac{1}{2}a{t^2}$
$ \Rightarrow {S_1} = {S_2}$
Substituting the given values of initial velocity and acceleration of car A and B respectively.
$ \Rightarrow 40t + \left( {\dfrac{1}{2} \times 0 \times {t^2}} \right) = 0t + \dfrac{1}{2} \times 2 \times {t^2}$
$ \Rightarrow 40t = {t^2}$
$ \Rightarrow t = 40$
Hence, both the cars will meet each other after 40s.
Note:
Velocity gives us information about the rate of change of the position of the body, meaning how fast the position of the body is changing per unit time. In physics, the velocity is defined as the displacement divided by time where displacement is defined as the difference between the final and initial positions of the body. Further, when an object travels the same distance every second, then the object is said to be moving with constant velocity. This describes that the magnitude of the velocity or speed and the direction of the velocity both remain constant.
Formula Used: $S = ut + \dfrac{1}{2}a{t^2}$
Complete answer:
The second equation of motion also known as the position time relation is mathematically given as, $S = ut + \dfrac{1}{2}a{t^2}$
Where ‘s’ is the displacement, ‘u’ is the initial velocity, ‘t’ is the time and ‘a’ is the acceleration of the body.
This formula is used to calculate the displacement when the values for time, acceleration and initial velocity are known.
Now,
Let the distance covered by car A be ${S_1}$
Let the distance covered by the car B be ${S_2}$
Both cars will meet each other when ${S_1} = {S_2}$
The position time relation of car A can be given as,
${S_1} = ut + \dfrac{1}{2}a{t^2}$
The position time relation of car B can be given as,
${S_2} = ut + \dfrac{1}{2}a{t^2}$
$ \Rightarrow {S_1} = {S_2}$
Substituting the given values of initial velocity and acceleration of car A and B respectively.
$ \Rightarrow 40t + \left( {\dfrac{1}{2} \times 0 \times {t^2}} \right) = 0t + \dfrac{1}{2} \times 2 \times {t^2}$
$ \Rightarrow 40t = {t^2}$
$ \Rightarrow t = 40$
Hence, both the cars will meet each other after 40s.
Note:
Velocity gives us information about the rate of change of the position of the body, meaning how fast the position of the body is changing per unit time. In physics, the velocity is defined as the displacement divided by time where displacement is defined as the difference between the final and initial positions of the body. Further, when an object travels the same distance every second, then the object is said to be moving with constant velocity. This describes that the magnitude of the velocity or speed and the direction of the velocity both remain constant.
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