Question

A particle A is moving along x-axis at time t-0, it has velocity of 10m/s and acceleration -4m/s2.Particle B has velocity of 20m/s and acceleration -2m/s2.Initially both particles are at the origin. At time t-2s, distance between the two particles is-
A) 20m. B) 24m
C) 36m. D) 42m

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Seperation\:between\:two\:particles=24\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Initial \: velocity(u_{1}) = 10 \: m/s \\ \\ \tt:\implies Acceleration( a_{1}) = - 4 \: m/{s}^{2} \\ \\ \tt: \implies Initial \: velocity(u_{2}) = 20 \: m/s \\ \\ \tt:\implies Acceleration( a_{2}) = - 2\: m/{s}^{2} \\ \\ \tt: \implies Time (t) = 2 \: sec \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Distance \: between \: them =?$

• According to given question :

$\bold{For \: particle \: A: } \\ \tt: \implies s_{1} = u_{1}t + \frac{1}{2} { a_{1}t}^{2} \\ \\ \tt: \implies s_{1} =10 \times 2 + \frac{1}{2} \times ( - 4) \times {2}^{2} \\ \\ \tt: \implies s_{1} =20 - 8 \\ \\ \green{\tt: \implies s_{1} =12 \: m} \\ \\ \bold{For \: particle \: B : } \\ \tt: \implies s_{2} = u_{2}t + \frac{1}{2} { a_{2}t}^{2} \\ \\ \tt: \implies s_{2} =20 \times 2 + \frac{1}{2} ( - 2) \times {2}^{2} \\ \\ \tt: \implies s_{2} =40 - 4 \\ \\ \green{\tt: \implies s_{2} =36 \: m} \\ \\ \bold{For \:Distance \: between \: them : } \\ \tt: \implies s= s_{2} - s_{1} \\ \\ \tt: \implies s=36 - 12 \\ \\ \green{\tt: \implies s=24 \: m}$

Answer 2

GIVEN :-

◙ Initial velocity of particle A (u) = 10 m/s

◙ Acceleration of particle A (a) = -4 m/s²

◙ Initial velocity of particle B (u') = 20 m/s

◙ Acceleration of particle B (a') = -2 m/s²

◙ Time(t) = 2 s

TO FIND :-

The distance between the particles.

FORMULA TO BE USED :-

$\sf{s=ut+\dfrac{1}{2}at^2}$

SOLUTION :-

For particle A,

$\sf{s=ut+\dfrac{1}{2}at^2 }\\\\\sf{\implies s=10(2)+\dfrac{1}{2}\times (-4)(2)^2 }\\\\\sf{\implies s=20+\dfrac{-16}{2} }\\\\\sf{\implies s=20-8}\\\\\boxed{\sf{\implies s=12\ m}}$

$\rule{200}{2}$

For particle B,

$\sf{\implies s'=u't+\dfrac{1}{2}a't^2}\\\\\sf{\implies s'=20(2)+\dfrac{1}{2}\times (-2)(2)^2 }\\\\\sf{\implies s'=40+(-4)}\\\\\boxed{\sf{\implies s'=36\ m}}$

Now, distance between particle A and particle B = s' - s

⇒Distance between particle A and particle B = 36 - 12

⇒Distance between particle A and particle B = 24 m

Therefore, the answer is (B) 24 m.