Question

A body starts from rest and moving with uniform acceleration of 4 m/s^2 covers half of its total path

during the last second of its motion. Find the time taken and the total distance travelled.

Class XI
Motion in a straight line
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No spams!! No cheating ¯\_(ツ)_/¯​

Answer 1

Let the total distance travelled by the particle be = x m

Let total time taken be = n s

Given that, a = 4 m/s^2

Initial velocity = 0 m/s

We know that, distance covered by a prticle in its nth second is:

$\sf{x_{n} = v_{0} + \frac{a}{2}(2n - 1)}$

Plug in the values:-

x/2 = 0 + 4/2(2n - 1)

x/2 = 2(2n - 1)

x/4 = 2n - 1

(x + 4)/8 = n

Now, we have both distance and time in terms of x, so we can plug in the values in the second equation of motion:

$\sf{x = v_{0}t + \frac{1}{2} at^{2}}$

=> x/2 = 0 + 1/2(4)((x+4)/8)^2

=> x = 4 × (x^2 + 8x + 16)/64

=> 16x = x^2 + 8x + 16

=> x^2 - 8x + 16 = 0

On solving, you get the equal roots as 4

Thus:

x = 4 m

t = n = (x + 4)/8 = (8)/8

t = 1 s