Question

A particle is moving in x-y plane. At time t = 0, particle is at (1m, 2m) and has velocity $(4\hat{i}+6\hat{j})$ m/s, At t = 4 s, particle reaches at (6m, 4m) and has velocity $(2\hat{i}+10\hat{j})$ m/s. In the given time interval, find.
(a) average velocity,
(b) average acceleration and
(c) from the given data, can you find average speed?

Answer 1

Total displacement $\left[\right(6-1{)}^2+(4-2{)}^2{]}^{\frac{1}{2}}$

=$\sqrt{29}m$

total time=$4$sec

change in velocity=|$(\widehat{2i}+\widehat{10j})-(\widehat{4i}+\widehat{6j})|$

$\sqrt{20}$

So Average velocity=total displacement /total time

=$\frac{\sqrt{2}9}{4}$ m/sec

Average acceleration=change in velocity/time interval

=$\frac{\sqrt{2}0}{4}=\frac{\sqrt{5}}{2}$m/sec

Since path of the particle is not know only initial and final position is known so average speed can not be calculated with this given data.