particle at rest starts at point O in staright line to point A. velocity is V, T seconds after l;eaving O?

Question

v=0.3t - 0.003t^2

a)find distance OA
b) show that for the motion from O to A max speed is 1.5 times greater than its average speed

Answer 1

I am assuming that the velocity of our particle is 0 once it reaches point O.

If we integrate v(t) we will get p(t) which gives us our distance at t seconds.

∫ 0.3t - 0.003t^2
= 0.15t^2 - 0.001t^3 + C
initial conditions claim that p(0) = 0, so C must equal 0 and
p(t) = 0.15t^2 -0.001t^3

Now if we solve v(t) for when it is 0, we will have either 0, since we start at rest, or some other time in seconds at which we stop.

v(t) = 0 = 0.3t - 0.003t^2
0 = t(0.3 - 0.003t)
*{t = 0
{0 = 0.3 -0.003t
{0.003t = 0.3
{t = 0.3/0.003
*{t = 100

So after 100 seconds, we have stopped at point O.

p(100) = 0.15*10,000 -0.001*1,000,000
p(100) = 1500 - 1000
p(100) = 500

The average speed is
(1/100)∫[from x = 0 to x = 100] v(t) dt
= (1/100)p(t)]{0:100}
=500/100
= 5

We are asked to show that the max speed is 1.5 times greater than the average, or 7.5

To find the max speed, I am going to differentiate our v(t) and seek out extrema.

v΄(t) = Dt v(t)
= Dt [0.3t - 0.003t^2]
= 0.3 - 0.006t

v΄(t) is defined on all real numbers, so our only critical number is when v΄(t) = 0, let's solve for it.

v΄(t) = 0 = 0.3 - 0.006t
0.006t = 0.3
t = 0.3/0.006
t = 300/6 = 50

Let's take the second derivative to perform the second derivative test to determine if we have found a maximum or minimum:

v΄΄(t) = Dt v΄(t)
= Dt [0.3 - 0.006t]
= -0.006

Since v΄΄(t) is always negative, the graph is concave downward, and our extremum at t = 50 is an absolute maximum.

Lastly, if we find the speed at t = 50, it should equal 7.5, let's see:

v(50) = 0.3(50) - 0.003(50^2)
= 15 - 0.003(2500)
= 15 - 7.5
= 7.5

Our max speed is indeed 1.5 times greater than our average speed.

--charlie

Answer 2

v=0.3t - 0.003t^2
a)let S=distance OA,which
is the integral of v
S=0.15t^2-0.001t^3+C,
where C is a constant
but S=0 at t=0
hence,C=0
therefore,
S=0.15t^2-0.001t^3

b)a=dv/dt=0 when v is at max
a=dv/dt=0.3-0.006t=0
0.006t=0.15
t=50 at max speed
therefore max speed
vmax=0.3*50-0.003*50^2
=15-7.5
=7.5
when the particle reaches
point A, v=0
0.3t - 0.003t^2=0
300t-3t^2=0
t(300-3t)=0
t=0 or t=100
eliminate t=0
hence,particle takes 100 time
units to travel from O to A
therefore,average speed
S100=distance tavelled/100
=0.15*100^2
-0.001*100^3)/100
=500/100=5
but max speed =7.5
ratio =7.5/5=1.5
therefore,the motion from O
to A max speed is 1.5 times
greater than its average speed

i hope that this helps

Answer 3

The area under the velocity-time graph is displacement.

a) So integrate v to give displacement

s = 0.15t^2-0.001t^3

The distance OA is the displacement.

b) At the max speed the acceleration is 0

v' = a = 0.3 - 0.006t

When 0.3 - 0.006t = 0 and this is true when t = 50.

so max speed is 0.3*50-.003(50)^2 = 7.5.
(It clearly isn't a minimum since v(t=0) = 0 which is less than 7.5)

Average velocity is displacement / time
Average velocity = 0.15t^2-0.001t^3/ t

=0.15t-0.001t^2

when t = 50; 0.15(50)-0.001(50)^2 = 5

max speed = 7.5 and average speed is 5

and 7.5/5 = 1.5

So max speed is 1.5 times greater than the average speed.

Answer 4

Nothing better to do.