Question

two objects are moving along the same straight line. they cross a point A with an acceleration a, 2a and velocity 2u, u at time t=0. the distance moved by the object when one overtakes the other is??

Answer 1

s = distance traveled
t = time for one object to overtake the other
s = u t +1/2 (2a) t^2 = 2u t + 1/2 a t^2
=> u t = 1/2 a t^2
So t = 2u/a
So. s = 6 u^2/a

Answer 2

Answer:

The distance moved by the object when one overtakes the other is $\bold{2 \mathrm{u}^{2} / \mathrm{a}}$

Step-by-step explanation:

Step 1:

Given Data:  

Let t be the time when the 2nd object overtake the 1st object

For the 1st object distance travelled will be:

Step 2:

$\mathrm{S}_{1}=\mathrm{ut}+1 / 2 \mathrm{at}^{2}$---1.

For the 2nd object distance travelled will be

Step 3:

\mathrm{S}_{2}=\mathrm{ut}+1 / 2 \mathrm{at}^{2}=> S2 = 2ut + at2 ---2.

Step 4:

Therefore at time t distance travelled by the 1st object must be equal to that travelled by the 2nd object:

S1 = S2  

$\begin{array}{l}{\mathrm{ut}+1 / 2 \mathrm{at}^{2}=2 \mathrm{ut}+\mathrm{at}^{2}} \\ {=>\mathrm{ut}-1 / 2 \mathrm{at}^{2}=0}\end{array}$

=> t(u- ½ at) = 0

=> t = 2u/a  

Step 5:

Now distance travelled when the bodies overtake will be

d = u× t = u×2u/a  

=$2 \mathrm{u}^{2} / \mathrm{a}$