Question

Check my answers. The position of a certain particle depends on time according to the following equation.
x(t) = t^2 - 5.4t +0.9 (x is meters, t is time in seconds)
a)find the displacement and average velocity for the interval: t1=3.4 and t2=4.1.

b)find the general formula for the displacement forthe time interval form t to t+delta t
delta x = ?

c)use the limiting process to obtain the instantenous velocity for any time t.
v(t) = ?


my answers (that are wrong) were:
a) 2.94m is displacement, (i have avg vel as correct at 2.1m/s)

b) delta x = v + delta t
c) v(t) = dx/dt.

Answer 1

a) The position at t1 = -5.9 m
The position at t2 = -4.43 m
Therefore the displacement between times t1 and t2 is 5.9 - 4.43 = 1.47 m

Answer 2

A displacement of 1.47 m in 0.7 s gives an average velocity of 1.47/0.7 = 2.1 m/s

Answer 3

ah i was doing 1.47*.7 not dividing. thank you. do you have any idea as to parts b and c? i am at a complete loss for that

Answer 4

\[\delta x=t ^{2}-5.4t+0.9-[(t+\delta t)^{2}-5.4(t+\delta t)+0.9]\]
\[=t ^{2}-5.4t+0.9-(t ^{2}+2t \delta t+\delta t ^{2}-5.4t-5.4 \delta t+0.9)\]
Therefore, simplifying:
\[\delta x=-2t \delta t-\delta t ^{2}+5.4 \delta t\]

Answer 5

c) The equation derived in part b) is
\[\delta x=-2t \delta t-\delta t ^{2}+5.4 \delta t\ ...........(1)\]
Taking out the common factor on the right hand side of (1) gives:
\[\delta x=\delta t(-2t-\delta t+5.4)\ ..........(2)\]
Dividing both sides of (2) by delta t we get:
\[\frac{\delta x}{\delta t}=-2t-\delta t+5.4\ ...........(3)\]
\[\lim_{\delta t \rightarrow 0}\frac{dx}{dt}=-2t+5.4\]