• Apr 14th 2007, 02:12 PM
bobchiba
1) Car is uniformely accelerated at 0.7m/sec^2 from rest, for 60sec, travels at constant velocity for 5 mins and comes to rest after uniformaly retarding for 20sec

- total distance traveled?
- retardation?
- number or revolutions of a wheel (diameter 0.5m)?
- max angular acceleration of wheel?

any ideas much apreciated!
• Apr 14th 2007, 04:09 PM
majik1213
1) Car is uniformely accelerated at 0.7m/sec^2 from rest, for 60sec, travels at constant velocity for 5 mins and comes to rest after uniformaly retarding for 20sec

- total distance traveled?

*break it up into 3 parts: t in seconds
1. 0<t<60 <- x=.5*a*t^2, t=60
2. 60<t<360 <- velocity constant => x=vt
3. 360<t<420 <- x=360v-0.5*a*3600

add em all up .. figuring out why i chose what i did along the way (kinematic equations are simple enough that I leave comprehension to you)

that should get you started

any ideas much apreciated!
• Apr 14th 2007, 04:17 PM
topsquark
Quote:

Originally Posted by bobchiba
1) Car is uniformely accelerated at 0.7m/sec^2 from rest, for 60sec, travels at constant velocity for 5 mins and comes to rest after uniformaly retarding for 20sec

- total distance traveled?
- retardation?
- number or revolutions of a wheel (diameter 0.5m)?
- max angular acceleration of wheel?

any ideas much apreciated!

"Retardation?" Is that supposed to mean "deceleration?" You need to be clearer. I'll assume it means constant deceleration and proceed from there.

The motion is in three phases:
1) Constant acceleration from rest for 60 s with a = 0.7 m/s^2.
2) Constant velocity for 5 min = 300 s.
3) Constant deceleration for 20 s, at which point it comes to rest.

Total distance travelled
Call the origin to be where the car started to accelerate and the +x direction to be the direction of motion. Then:
Phase 1
x1 = x0 + v0*t + (1/2)a*t^2 <-- x0 = 0 m, v0 = 0 m/s
x1 = (1/2)a*t^2
x1 = (1/2)*0.7*(60)^2 m = 1260 m

v = v0 + a*t <-- v0 = 0 m/s (We'll need v after 60 s for the next two parts.)
v = a*t
v = 0.7*60 m/s = 42 m/s

Phase 2
x2 = v*t
x2 = 42*300 m = 12600 m

Phase 3
We don't know a. I'm going to reset v to be v0 in the equations. Final v = 0 m/s.
x3 = (1/2)*(v0 + v)*t
x3 = (1/2)*(42 + 0)*20 m = 420 m

So the total distance is
x = x1 + x2 + x3 = 1260 m + 12600 m + 420 m = 14280 m

The deceleration from phase 3 may be found by
v = v0 + a*t
0 = 42 + 20a
a = -42/20 m/s^2 = -2.1 m/s^2
(The negative sign indicates that the acceleration here is in the -x direction, which is opposite to the direction of the velocity. Hence the acceleration is, in fact, a deceleration.)

Revolutions
All we need here is the total distance travelled and the radius of the wheel. If the wheel is rolling without slipping (a standard assumption unless otherwise mentioned) the relationship is:
x = r*(theta)
where x is the total distance travelled, r is the radius of the wheel, and (theta) is the magnitude of the angular displacement (in radians.)

So
(theta) = x/r = 14280 m/0.25 m = 57120 rad

57120 rad = 57120/(2(pi)) rev = 9090.93 rev

Angular acceleration
The tire does not have any angular acceleration during phase 2, so all we need worry about is the angular accelerations from phases 1 and 3. The angular acceleration (alpha) of the tire is related to the translational acceleration (a) by:
a = r*(alpha)
So the larger the acceleration the larger the angular acceleration. Thus the largest angular acceleration is in phase 3, with a = -2.1 m/s^2. (Drop the negative sign since all we care about here is the magnitude.)

So:
(alpha) = a/r = (2.1 m/s^2)/(0.25 m) = 8.4 rad/s^2

-Dan
• Apr 14th 2007, 05:59 PM
majik1213
to be certain, deceleration means a decrease in the magnitude of acceleration. In otherwords, deceleration can never take on negative values. In contrast, retardation likely refers to negative ACCeleration; that is, acceleration that causes the particle to come to a stop.