AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

The acceleration of the body is calculated as
a=‌

F

m

a=‌

(2.6 ∧ i +1.6 ∧ j )

2

a=(1.3

∧

i

+0.8

∧

j

)

m

s2

The velocity vector is,
v=∫adt
v=∫(1.3

∧

i

+0.8

∧

j

)dt
v=(1.3

∧

i

+0.8

∧

j

)t+c
Apply the given condition in above expression, i.e. at t=0
v=(3.6

∧

i

−4.8

∧

j

)

m

s

(3.6

∧

i

−4.8

∧

j

)‌‌=(1.3

∧

i

+0.8

∧

j

)(0)+c
c‌‌=(3.6

∧

i

−4.8

∧

j

)
Thus, the velocity is,
v=(1.3

∧

i

+0.8

∧

j

)t+(3.6

∧

i

−4.8

∧

j

)
v=(1.3t+3.6)

∧

i

+(0.8t−4.8)

∧

j

According to question, vy=0
Thus,
0.8t−4.8‌‌=0
t‌‌=6s