# Thread: HW typo--

1. ## HW typo--

A new motion detector is used to graph the position of a sports car. If the themotion detector is not callibrated properly, it might give unrealistic graph data. The position of the car is described by the function S= 2t^3-21t^2 + 60t, tE [0,6], where s is measured in meters and t in seconds. it is known the largest acceleration is 12 m/s ^2.

Can some one double check this ? and help me

a) WHat is its initial velocity? ---60
b)When is the car at rest? t=0
c)When is the car moving in the forward direction? ******* Can someone explain to me how to get this?
d)draw a diagram ----
e)Find the total distance traveled by the car in the first 6 seconds----36
f)Find the max acceleration of the car? a = 12t-42 a=30

2. Originally Posted by tnkfub
a) WHat is its initial velocity? ---60
b)When is the car at rest? t=0
c)When is the car moving in the forward direction? ******* Can someone explain to me how to get this?
d)draw a diagram ----
e)Find the total distance traveled by the car in the first 6 seconds----36
f)Find the max acceleration of the car? a = 12t-42 a=30
Well, a is correct but how can b be correct then? If the initial speed (at t = 0) is 60m/s, then the car can't be at rest at t = 0. The car is at rest when v = 0, take it from there.
For c, when v is positive: the car moves forward (and backward when v < 0).
Draw the diagram yourself, it'll help for e as well, f seems to be ok.

3. thanks for the help TD

b)

S= 2t^3-21t^2 + 60t
v = 6t^2-42t+60
v= 6(t^2-7t+10)
v= 6(t-2)(t-5)
t= 2 and 5 sec

does that make sense that v is 0 at two points of time?

This doesnt look right >> :/

4. Originally Posted by tnkfub
thanks for the help TD

b)

S= 2t^3-21t^2 + 60t
v = 6t^2-42t+60
v= 6(t^2-7t+10)
v= 6(t-2)(t-5)
t= 2 and 5 sec

does that make sense that v is 0 at two points of time?

This doesnt look right >> :/
Why not? If you made a good diagram, you can see why.