TG EAMCET 3 May 2025 Shift 1 Paper

Distance covered by the particle in first 3 seconds.

s‌=ut+‌

1

2

at2
‌=0×3+‌

1

2

×2×32
‌=9m
Velocity of particle after 3 s .

v=u+at=0+2×3=6m∕s
After t=3s, direction of acceleration is reversed, so new acceleration,

a′=−2m∕s2
Since, particle returns to its initial position, thus,

‌s1+s2=0
‌s2=−s1=−s=−9m
∴‌‌s2=vt′+‌

1

2

a′(t′)2
‌−9=6t′+‌

1

2

×(−2)(t′)2
⇒‌‌(t′)2−6t′−9=0
⇒‌‌t′=‌

−(−6)±√36+36

2×1

‌=3±3√2
⇒‌‌t′=(3+3√2)s
The total time from the beginning of the motion until the particle returns to its initial position,

‌T=3+t′
‌=3+3+3√2=6+3√2
‌=3(2+√2)s