Question

Two bodies start with the same velocity but one move with acceleration 2m/s^2 and the other with 3m/s^2.After 10 sec,the second body is ahead of the first body by ____m.

Answer 1

◙ Given ◙

Two bodies start with the same velocity.

• Initial velocity of body A (u) and initial velocity of body B (U) = 0 m/s
• Acceleration of body A (a) = 2 m/s²
• Acceleration of body B (A) = 3 m/s²

◙ To Find ◙

After time, t = 10 s, by how much distance will the second body be far from the first body.

◙ We Must Know ◙

$\boxed{\bigstar\ \sf{\color{red}{v=u+at}}}$

$\boxed{\bigstar\ \sf{\color{red}{v^2=u^2+2as}}}$

Where,

• v is the final velocity
• u is the initial velocity
• a is the acceleration
• t is the time taken
• s is the distance covered

◙ Solution ◙

◘ For body A :-

u = 0 m/s

a = 2 m/s²

t = 10 s

Putting the values :-

$\sf{\longrightarrow v=0+2(10)}\\\\\sf{\longrightarrow v=20\ m/s}$

$\rule{90}1$

$\sf{v^2=u^2+2as }\\\\\sf{\longrightarrow (20)^2=(0)^2+2(2)s}\\\\\sf{\longrightarrow 400=4s}\\\\\sf{\color{lime}{\longrightarrow s=100\ m}}$

$\rule{180}2$

◘ For body B :-

U = 0 m/s

A = 3 m/s²

T = 10 s

Putting the values :-

$\sf{\longrightarrow V=0+3(10)}\\\\\sf{\longrightarrow V=30\ m/s}$

$\rule{90}1$

$\sf{V^2=U^2+2AS}\\\\\sf{\longrightarrow (30)^2=(0)^2+2(3)S}\\\\\sf{\longrightarrow 900=6S}\\\\\sf{\color{lime}{\longrightarrow S=150\ m}}$

Distance between body A and body B = S - s

= 150 - 100

= 50 m

Hence, body B is 50 m ahead of body A.

Answer 2

Given:

• Two bodies start with same velocity.
• Acceleration of one body is 2m/s²
• And other body by 3m/s²

To Find :

By how much distance ,second body ahead of the first body ,after 10 sec

Solution :

Suppose both bodies Start with intital velocity v .

For body 1 :

Acceleration = 2m/s²

Initial Velocity=V

t= 10sec

By third equation of motion :

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

$\sf\:S_1=ut+\dfrac{1}{2}\times2\:t^2$

$\sf\:S_1=ut+t^2$

put t = 10 sec

$\sf\:S_1=10v+100...(1)$

For body 2:

Acceleration = 3m/s²

Initial Velocity=V

t= 10sec

By third equation of motion :

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

$\sf\:S_2=ut+\dfrac{1}{2}\times3\:t^2$

Put t = 10sec

$\sf\:S_1=10v+\dfrac{3}{2}\times100...(2)$

Therefore,

The second body is ahead of the first body after 10 sec

$\sf=S_2-S_1$

$\sf=(10v+\dfrac{3}{2}\times\:100)-(10v+100)$

$\sf=50m$

$\rule{200}2$

${\purple{\boxed{\large{\bold{Formula's}}}}}$

Kinematic equations for uniformly accelerated motion .

$\bf\:v=u+at$

$\bf\:s=ut+\frac{1}{2}at{}^{2}$

$\bf\:v{}^{2}=u{}^{2}+2as$

and $\bf\:s_{nth}=u+\frac{a}{2}(2n-1)$