Question

a particle moves with constant acceleration along a straight line starting from rest. the percentage increase in its displacement during the 4th second compared to that in the 3rd second is? (ans : 40%)

Answer 1

⛦Hᴇʀᴇ Is Yoᴜʀ Aɴsᴡᴇʀ⚑➧➾
▬▬▬▬▬▬▬▬▬▬▬▬☟

➧ We know that
➾ Sn th = u + 12a(2n − 1)

➾ S 3ʳᵈ = 0 + 12a (2 × 3 − 1)
➾ 52a (forn = 3s)

➾ S 4ᵗʰ = 0 + 12a (2 × 4 − 1)
➾ 72a (forn = 4s)

➧ So, the percentage increase
S 4ᵗʰ

➾ S3ʳᵈ S3ʳᵈ × 100
➾ 72a − 52a 52a × 100

➾ 2a 252a × 100
➾ 2 × 20 = 40 ...✔

_________
Thanks...✊

Answer 2

Answer:

The percentage % increase in the displacement of particle, during the 4th second compared to that in the 3rd second is equal to 40%.

Explanation:

We know that the displacement of a particle of $n^{th}$ second of a uniformly accelerated body is given by:

$S_{n }=u+\frac{1}{2}a(2n-1)$                                                             ...............(1)

The particle was at rest initially, u=0

Displacement for 3rd second will be:

Substitute the value of 'u' and n = 3 in the equation (1);

$S_{3}=0+ \frac{1}{2} a(2(3)-1)$

$S_{3}=\frac{5}{2}a$

Displacement for 4th second will be:

$S_{4}=0+ \frac{1}{2} a(2(4)-1)$

$S_{4}=\frac{7}{2}a$

The percentage increase in the displacement :

% increase displacement $=\frac{S_{4}-S_{3}}{S_{3}}*100$

                                          $=\frac{\frac{7}{2}a -\frac{5}{2} a}{\frac{5}{2}a } *100$

                                          $=\frac{2a/2}{5a/2}*100$

                                          $=40\%$

Therefore, when displacement during the 4th second compared to that in the 3rd second, percentage increase is equal to 40%.