Question

A body has initial velocity 3 m/s along aA and it is subjected to an acceleration 2 m s-2 perpendicular to CA. The magnitude of displacement of body in
time interval from t= 0 to t=4 second will be
B
at
Thing
a
O 12 m
O 20 m
O 16 m
28 m​

Answer 1

Answer:

Velocity, V

OE

=3 m/s

Since, applied force is perpendicular to OE so the velocity V

OE

will be constant.

So, displacement along OE in 4 sec S

OE

=3×4=12 m

Along OF

Force applied, F=4 N

Mass of the body, m=2 kg

So, acceleration a=

m

F

=2 m/s

2

Displacement along OF in time t=4 sec

S

OF

=ut+

2

1

at

2

S

OF

=0+

2

1

×2×4

2

S

OF

=16 m

After t=4 sec, the distance of the body from O to OP

OP

2

=12

2

+16

2

OP

2

=144+256

OP

2

=400

OP=20 m