Question

A car starting from rest and moving along
a straight path with uniform acceleration
covers distances p and q
in the first two
successive equal intervals of time. Find
the ratio of p to q.​

Answer 1

Answer:

Let the time intervals be 't' seconds each, and the distances covered after t seconds and 2t be x1 and x2 respectively. ( Dont take the 1 and 2 as powers, they are subscripts here)

Then, from the question, p=x1 and q=x2-x1.

Then from the 2nd equation of motion, since intial velocity is 0(ie. u=0), so

x1= (1/2)∗at2 = 1/2 a(t^2)

And, x2= (1/2)∗a(2t)2 = 2a(t^2)

Where a is the constant acceleration.

So, p=1/2 a(t^2)

And q=2a(t^2)-1/2a(t^2) = 3/2a(t^2)

So p/q=1/3

Answer 2

$\huge\mathbb\pink{Question:-}$

A car starting from rest and moving along a straight path with uniform acceleration

covers distances p and q in the first two successive equal intervals of time. Find

the ratio of p to q.

$\huge\mathbb\blue{Answer:-}$

Let the time intervals be 't' seconds each, and the distances covered after t seconds and 2t be x1 and x2 respectively. ( Dont take the 1 and 2 as powers, they are subscripts here)

Then, from the question, p=x1 and q=x2-x1.

Then from the 2nd equation of motion, since intial velocity is 0(ie. u=0), so

x1= (1/2)∗at2 = 1/2 a(t^2)

And, x2= (1/2)∗a(2t)2 = 2a(t^2)

Where a is the constant acceleration.

So, p=1/2 a(t^2)

And q=2a(t^2)-1/2a(t^2) = 3/2a(t^2)

So p/q=1/3

Special note- This is quite interesting, if you try to find the ratios for successive intervals they will be 1:3:5:7…, and this ratio is always the same, in all cases, where acceleration is constant, for example ratio of distance travelled by a body in free fall is also 1:3:5:7:9:11… (till it doesn't hit the ground)