Question

V= 2i+4j at time t=0. A= I+3j for 4s. (I) find coordinates of the object which is at origin at t=0. (II) Find magnitude of velocity at the end of 4s.

Answer 1

Answer:

(i) Here initial position of the object

r0=x0iˆ+y0jˆ+0iˆ+0jˆ

Initial velocity, v0=v0xiˆ+v0yjˆ=2iˆ+4jˆ

Acceleration a=axiˆ+ayjˆ=iˆ−3kˆ

And t=4s

Let the final coordinates of the object be (x,y),. Then according to the equation for the path of particle under constant acceleration.

x=x0+v0xt+12axt2

=0+2×4+12(1)×42

⇒x=16cm

and y=y0+v0yt+12ayt2

=0+4×4+12(−3)×42

→y=−8m

Therefore the object lies at (16iˆ−8jˆ) at t=4s.

(ii) Using equation

v=v0+at

⇒v=(2iˆ+4jˆ)+(iˆ−3jˆ)×4

=(2iˆ+4jˆ)+(4iˆ−12jˆ)=(2+4)iˆ+(4−12)jˆ

⇒v=6iˆ−8jˆ

∴ Magnitude of velocity, |v|=62+82−−−−−−√=10ms−1

Its direction with X-aixs θ=tan−1(−86)≈−53∘

Explanation:

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