
Two particles start moving from the same point along the same straight line. The first moves with constant velocity v and the second with constant acceleration a. During the time that elapses before the second catches the first, the greater distance between the particles is ?
A. $\dfrac{{{v}^{2}}}{a} \\ $
B. $\dfrac{{{v}^{2}}}{2a} \\ $
C. $\dfrac{2{{v}^{2}}}{a} \\ $
D. $\dfrac{{{v}^{2}}}{3a}$
Answer
163.8k+ views
Hint:Here we use Newton's equation of motion to solve this problem. We know that distance between two particles is maximum when they have the same velocity. Here everything is given. We have to just formulate the equation and apply extreme conditions to find maximum distance.
Formula used:
Here mainly Newton's second equation of motion is used.
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
where t= time, s= distance travelled, u=initial velocity and a= acceleration.
Complete step by step solution:
A particle is said to have constant velocity, when it covers equal distances in equal intervals of time. A particle is said to have constant acceleration when the particle covers equal velocity in equal intervals of time. If a body has constant velocity, then we can also say that the body is in uniform motion.
Here both particles start from the same point and they are moving in a straight line. But the first body is moving with constant velocity and the second one is moving with constant acceleration. We have to find the maximum distance between both of them before the second catches the first.
Let $x$ be the distance between the particles after t seconds. Form newton’s second equation of motion we can write that:
$x=vt-\dfrac{1}{2}a{{t}^{2}}$
For x to be maximum,
$\dfrac{dx}{dt}=0$
Therefore, on differentiating and equating it to zero we get time, $t=\dfrac{v}{a}$.
On substituting the value of time in the equation of distance, we get greater distance between the particles. That is,
$x=v\left( \dfrac{v}{a} \right)-\dfrac{1}{2}a{{\left( \dfrac{v}{a} \right)}^{2}} \\
\therefore x=\dfrac{{{v}^{2}}}{2a}$
Therefore, the correct answer is option B.
Note: Generally, in questions asking minimum and maximum parameters we have to formulate an appropriate equation and apply extreme conditions we have studied in differentiation to solve the problem. The distance between two particles is greater only when both particles are travelling with the same velocity.
Formula used:
Here mainly Newton's second equation of motion is used.
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
where t= time, s= distance travelled, u=initial velocity and a= acceleration.
Complete step by step solution:
A particle is said to have constant velocity, when it covers equal distances in equal intervals of time. A particle is said to have constant acceleration when the particle covers equal velocity in equal intervals of time. If a body has constant velocity, then we can also say that the body is in uniform motion.
Here both particles start from the same point and they are moving in a straight line. But the first body is moving with constant velocity and the second one is moving with constant acceleration. We have to find the maximum distance between both of them before the second catches the first.
Let $x$ be the distance between the particles after t seconds. Form newton’s second equation of motion we can write that:
$x=vt-\dfrac{1}{2}a{{t}^{2}}$
For x to be maximum,
$\dfrac{dx}{dt}=0$
Therefore, on differentiating and equating it to zero we get time, $t=\dfrac{v}{a}$.
On substituting the value of time in the equation of distance, we get greater distance between the particles. That is,
$x=v\left( \dfrac{v}{a} \right)-\dfrac{1}{2}a{{\left( \dfrac{v}{a} \right)}^{2}} \\
\therefore x=\dfrac{{{v}^{2}}}{2a}$
Therefore, the correct answer is option B.
Note: Generally, in questions asking minimum and maximum parameters we have to formulate an appropriate equation and apply extreme conditions we have studied in differentiation to solve the problem. The distance between two particles is greater only when both particles are travelling with the same velocity.
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