TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Given, acceleration of particle,
a=(4

∧

i

+4‌

∧

j

)‌ms−2
Let at time, t=0
Velocity, ux=4

∧

i

ms−1
and uy=0

∧

j

ms−1
Speed of body be vx along X-axis and vy along Y-axis after covering a displacement sx of 6m.
Now, as we know that,
v2−u2=2as
and for displacement along X-axis
vx2−ux2=2axsx
⇒‌vx2−16‌=2×4×6
⇒‌vx‌=√48+16
⇒‌vx‌=√64=8ms−1
‌∴‌vx‌=8

∧

i

ms−1
Now, Since, ‌‌v=u+at
∴‌‌t=‌

vx−ux

ax

where, t is time taken. ∴‌‌t=‌

8−4

4

=2s
Again ‌‌vy=uy+ayt
∴‌vy‌‌=0+4×1
‌‌‌=4ms−1
⇒‌vy‌‌=4

∧

j

ms−1
‌∴‌v‌‌=vx

∧

i

+vy

∧

j

=8

∧

i

+4

∧

j

Now, resultant vector
|v|‌‌=√vx2+vy2
‌‌=√82+42
‌‌=√64+16
‌‌=√80=4√5ms−1