Question

-
A particle moves in the xy plane, starting
from the origin at t = 0 with an initial
velocity having an x component of
20 ms-' and a y component of - 15 ms-l. The
particle experiences an acceleration in the
x direction, given by a = 4 ms-2. Determine
the total velocity vector at any time.
(A) V, = (10 +6+) î- 150
(B) = (20 + 4t) î–150 Î
(C) 7= (20+41) î–15
(D) 1x = (5+6t) î– 15 h​

Answer 1

Answer:

v=[(20+4.0t)^x−15^y] m/s

Explanation:

In constant accelerated motion, we have the following relations:

1. v=u+at

(Velocity-Time Relation)

2. S=ut+12at2

(Position- Time Relation)

3. v2=u2+2aS

(Position- Velocity Relation)

Where,

U is the initial velocity,

V is the final velocity,

S is the displacement,

a is the acceleration and

t is the time.

Given:

u=(20^x−15^y) m/s

a=(4.0^x) m/s2

a)

Using the velocity-time relation, the velocity vector at any time is,

v=u+at=(20^x−15^y) m/s+(4.0t^x) m/s2

v=[(20+4.0t)^x−15^y] m/s