Question

A small frictionless cart, attached to the wall by a spring, is pulled 10 cm back from its rest position
and released at time t = 0 to roll back and forth for 4 sec. Its position at time t is
s = 1 - 10 cos piΔt. What is the cart's maximum speed? When is the cart moving that fast? What is the
magnitude of of the acceleration then?

Answer 1

as you may have learned in physics, the time derivative of position is velocity.

s'(t) = v(t) = 10pi*sin(pi*t)). this is maximum when |sin(pi*t)| = 1 since sin can only be between -1 and 1.

from calculus, the function s(t) has maximum values when s'(t) = 0. must solve s'(t) = 0 for t.

remember that sin is periodic meaning that it will return to that maximum speed every (pi or 2pi depending how you wanna look at it) radians. you must pick a t within 0 and 4 [seconds]

s''(t) = acceleration (time derivative of velocity)

Answer 2

ok what next

Answer 3

well its pretty much done. all tat is left is to solve

Answer 4

max speed = 10pi * 1 = 10pi cm/s

time of max is t = 1/2 + or - 2kpi . k is any real integer. only between t = 0 and 4 for k = 0

v is max as t = 0.5

a(t) = -10(pi^2)cos(pi*t)

a(0.5) = 0 cm/s^2 which is expected.

the spring reaches maximum velocity at maximum stretch, then it turns around. for velocity to change direction, acceleration must have been 0 at some point.

Answer 5

does this make sense to you?
Speed is ds/dt = 10πsin(πt) and its maximum value is 10π cms^-1
This is when sin(πt)=1 which is when t=1/2, 5/2, seconds
The acceleration is d^s/dt^2=10π²cos(πt)
When t=1/2, acceleration =0 since cos(π/2)=0,
t=5/2, acceleration =0 since cos(5π/2)=0

Answer 6

yes

Answer 7

doesnt to me because i cant match it

Answer 8

accel would be negative though

Answer 9

-10pi^2cos

Answer 10

its 1/2 and 3/2

Answer 11

A) 10pi=􀎔 􀁙 31.42 cm/sec; t = 0.5 sec, 2.5 sec; acceleration is 1 cm/sec^2
B) 10pi=􀎔 􀁙 31.42 cm/sec; t = 0 sec, 1 sec, 2 sec, 3 sec; acceleration is 0 cm/sec^2
C) pi=􀎔 􀁙 3.14 cm/sec; t = 0.5 sec, 1.5 sec, 2.5 sec, 3.5 sec; acceleration is 0 cm/sec^2
D) 10pi=􀎔 􀁙 31.42 cm/sec; t = 0.5 sec, 1.5 sec, 2.5 sec, 3.5 sec; acceleration is 0 cm/sec^2

Answer 12

.5 and 2.5 but none of them match to this with 0cmsec

Answer 13

1/2, 3/2, 5/2 and 7/2

Answer 14

D

Answer 15

3.14 cm sec? not 31.42?

Answer 16

it is D. the weird squares lol

Answer 17

are u super sure?

Answer 18

but what is with the extra values other than .5 and 2.5

Answer 19

yes. |v| is 10pi at sin((pi/2 + kpi)t) k is an real integer.

(pi/2 + kpi) is between 0 and 4 for k = 0 , 1, 2, 3

Answer 20

meaning that the values can be sin(pi/2), sin(3pi2/2), sin(5pi/2), sin(7pi/2).

1/2 to 7/2 is between 0 and 4. and all 4 of these = 1 or - -1 which is the max values of v

Answer 21

1 or -1*

Answer 22

so its d?

Answer 23

yes

Answer 24

thanks