Question

An object moves along the x axis according to the equation
x = 3.50t2 − 2.00t + 3.00,
where x is in meters and t is in second
Determine average acc/ average speed between 1.9s and 3.30s
find instantaneous speed / acc at t=1.9
Determine the instantaneous speed / acc at t = 3.30 s.
(e) At what time is the object at rest?

Answer 1

hi
you can use from this\[V=\Delta x/\Delta t\] for average v
& differentiate respect to t from x equation yields v equation if put t in it instantaneous v yields

Answer 2

for acc differentiate proportion to t from v equation and put t in it for instantaneous a

Answer 3

for average a use it\[a=\Delta V/\Delta t\]

Answer 4

when body recieve to rest it's velosity be zero so find time of that from velocity equation \[0=at+V _{0}\]

Answer 5

We have
\[x=3.5t^2-2.00 t+3.00\]
let's find velocity, we know that velocity is given as \(\frac{dx}{dt}\)
let's differentiate x with respect to t
\[\frac{dx}{dt}= 2*3.5t^{2-1}-2.00+0\]
so velocity v
\[v=7t-2\]

we have to find average speed between 1.9 s and 3.3 seconds
so
average speed =\( \frac{1}{3.3-1.9}\int_{1.9}^{3.3} (7t-2) dt \)
\[ \frac{1}{1.4}[ 7t^2/2-2t]_{1.9}^{3.3}\]
\[ \frac{1}{1.4}[(7*3.3^2/2-2*3.3)-(7*1.9^2/2-2*1.9)]\]
we get
\[\frac{1}{1.4} * 22.68\]= 16.2 m/s

Answer 6

for instantaneous speed at t=1.9 and 3.3 sec
substitute t=1.9 and 3.3 in
v=7t-2
for acceleration, differentiate v with respect to time
will get
a=7 m/s^2
so a is 7m/s^2 at t=1.9 sec and 3.3 sec