Question

A particle is travelling with velocity (4i+5j)m/s. It undergoes an acceleration of magnitude 2.5m/s^2 in a direction given by the vector (3i-4j). Find the velocity and displacement of the particle from its initial position after 4s.

Answer 1

In the direction of x
u = 4i, a= 2.5(3/5)
At t =4
v = u+at
v= 4 + 2.5 (3/5)4=4+6=10
In the direction of y
u = 5j, a= 2.5(-4/5)
At t =4
v = u+at
v= 5 + 2.5 (-4/5)4=5-8=-3
Hence
V = 10i-3j
Now we have initial and final velocities
r = 1/2(u +v)t
r = 1/2(4i+5j+10i-3j)4
r = 2(14i+2j)
r = 18i+4j