Question

A car starting from rest is moving with constant
acceleration in a straight line. The ratio of
displacement in first 2 second to the next
4 second is​

Answer 1

GiveN:-

• A car starting from rest is moving with constant acceleration in a straight line

To FinD :-

• The ratio of displacement in first 2 second to the next 4 second .

SolutioN :-

Here the initial Velocity of the car will be 0m/s. Let us take the acceleration of the car be a . So in first case using second equation of motion we have ,

C A S E - O N E :-

$\red{\bigstar}\underline{\textsf{ Using the second equation of motion :- }}$

$\sf\dashrightarrow\pink{ s = ut +\dfrac{1}{2}at^2 } \\\\\sf\dashrightarrow s_1 = (0ms^{-1} )(t)+\dfrac{1}{2}(a)(2)^2 \\\\\sf\dashrightarrow s_1 = \dfrac{4a}{2} \\\\\sf\dashrightarrow\underset{\blue{\sf Required\ Distance_{in \ 2 \ s } }}{\underbrace{\boxed{\pink{\frak{ Distance_1 = 2a }}}}}$

$\rule{200}2$

C A S E - T W O :-

Now for the next 4 seconds , the intial velocity can be find out using the first equation of motion that is v = u + at . On substituting the respective values we will get v that is Initial velocity in Second case as v = 0+2a = 2a . Now again using the second equation of motion ,

$\red{\bigstar}\underline{\textsf{ Using the second equation of motion :- }}$

$\sf\dashrightarrow\pink{ s = ut +\dfrac{1}{2}at^2 } \\\\\sf\dashrightarrow s_2 = (2a)(4) + \dfrac{1}{2}\times a \times (4)^2 \\\\\sf\dashrightarrow s_2 = 8a + 8a \\\\\sf\dashrightarrow\underset{\blue{\sf Required\ Distance_{in \ 4 \ s } }}{\underbrace{\boxed{\pink{\frak{ Distance_2 = 8a }}}}}$

Hence the required ratio will be equal to S1 :S2 = 2a : 8a = 1:4 .