Answer
77.7k+ views
Hint:Here Newton’s equation of motion is used. It is given that two particles start from the same position. They have the same acceleration but in opposite directions. The velocity of P when it overtakes Q is given. From this we can apply Newton's equation of motion to find out the velocity of Q at the point B.
Formula used:
Newton’s equations of motion are used.
1.$v=u+at$
2.$s=ut+\frac{1}{2}a{{t}^{2}}$
3.${{v}^{2}}={{u}^{2}}+2as$
where v= final velocity
u=initial velocity
a= acceleration
t= time
s= distance travelled
Complete step by step solution:
It is given that both particles simultaneously start from the same point but with different velocities.
Initial velocity of particle P, $v=15\,m{{s}^{-1}}$
Initial velocity of particle Q, \[v=20\,m{{s}^{-1}}\]
Let ‘a’ be the acceleration of particle P and ‘-a’ be the acceleration of particle Q. Then by applying Newton’s first equation of motion to particle P at the position B, we get:
Velocity, $30=15+at$
That is, $at=15$
Now consider velocity of particle Q at the position B be v. Now applying Newton’s first equation of motion to particle Q at the position B, we get:
Velocity of Q at B, \[v=20-at\]
On substituting, we get:
Velocity of Q at B,
\[v=20-15=5\,m{{s}^{-1}}\]
Hence, the answer is option B.
Note: We can easily solve it if we know how to apply Newton’s equation of motion. Don’t get confused with the concept of relative velocity. Negative acceleration is taken for Q because acceleration is a vector quantity and has both direction and magnitude and it is given that both particles have opposite acceleration.
Formula used:
Newton’s equations of motion are used.
1.$v=u+at$
2.$s=ut+\frac{1}{2}a{{t}^{2}}$
3.${{v}^{2}}={{u}^{2}}+2as$
where v= final velocity
u=initial velocity
a= acceleration
t= time
s= distance travelled
Complete step by step solution:
It is given that both particles simultaneously start from the same point but with different velocities.
Initial velocity of particle P, $v=15\,m{{s}^{-1}}$
Initial velocity of particle Q, \[v=20\,m{{s}^{-1}}\]
Let ‘a’ be the acceleration of particle P and ‘-a’ be the acceleration of particle Q. Then by applying Newton’s first equation of motion to particle P at the position B, we get:
Velocity, $30=15+at$
That is, $at=15$
Now consider velocity of particle Q at the position B be v. Now applying Newton’s first equation of motion to particle Q at the position B, we get:
Velocity of Q at B, \[v=20-at\]
On substituting, we get:
Velocity of Q at B,
\[v=20-15=5\,m{{s}^{-1}}\]
Hence, the answer is option B.
Note: We can easily solve it if we know how to apply Newton’s equation of motion. Don’t get confused with the concept of relative velocity. Negative acceleration is taken for Q because acceleration is a vector quantity and has both direction and magnitude and it is given that both particles have opposite acceleration.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write an article on the need and importance of sports class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Choose the one which best expresses the meaning of class 9 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)