Answer
3a) v ( t ) = $\frac{π}{2}$ cos ( $\frac{π t}{2}$ )
3b) v ( 1 ) = 0
3c) The particle is at rest when ( t ) = odd integer = 1 + 2n
3d) [ 0 , 1 ) ∪ ( 3 , 5 ) ∪ ( 7 , 9 ) …
3e) The distance traveled in the first 6 seconds is 6 feet.
3f) See attached sketch of 3f)
3g) a ( t ) = - $\frac{π²}{4}$ sin ( $\frac{π t}{2}$ )
3g) a ( 1 ) = - $\frac{π²}{4}$
3h) see attached photo of graph
3i) The particle is speeding up on the interval (1 , 2) ∪ (3 , 4) ∪ (5 , 6].
3i)The particle is slowing down on the interval [0 , 1) ∪ (2 , 3) ∪ (4 , 5).
Work Step by Step
3) f ( t ) = sin ( $\frac{π t}{2}$ )
a) Find the velocity at time ( t ).
Use Chain Rule: f ( x ) = g ( h ( x ) ) f ‘( x ) = d/dx [ g ( h ( x ) ] * d/dx [ h ( x ) ]
f ‘( t ) = v ( t ) = d/dx [ sin ( $\frac{π t}{2}$) ] * d/dx [ $\frac{π}{2}$ t ]
v ( t ) = ( cos ( $\frac{π t}{2}$ ) * ( $\frac{π}{2}$ )
3a) v ( t ) = $\frac{π}{2}$ cos ( $\frac{π t}{2}$ )
b) What is the velocity after 1 second?
v ( 1 ) = $\frac{π}{2}$ cos ( $\frac{π ( 1 )}{2}$ )
v ( 1 ) = $\frac{π}{2}$ cos ( $\frac{π}{2}$ )
cos ( $\frac{π}{2}$ ) = 0
v ( 1 ) = $\frac{π}{2}$ ( 0 )
3b) v ( 1 ) = 0
c) When is the particle at rest?
When is the velocity function equal to zero?
0 = $\frac{π}{2}$ cos ( $\frac{π t}{2}$ )
When is cosine function equal to zero?
cos t = 0
$cos^{- 1}$ cos t = $cos^{- 1}$ 0
t = $cos^{- 1}$ ( 0 ) ← Calculating in radian mode will output a difficult decimal, use degree mode for calculating inverse trig functions then convert to manageable radian fractions.
t = 90 degrees
Convert degrees to manageable radian fractions.
t = ( 90 degrees ) * ( $\frac{π}{180 degrees}$ )
t = $\frac{π}{2}$ ← The range of inverse cosine functions on calculators is [0,π] so be sure to include other values as they apply.
A graph of y = f ( t ) = cos t demostrates a pattern of t-values:
$\frac{π}{2}$ , $\frac{π}{2}$ + π , $\frac{π}{2}$ + π + π , … $\frac{π}{2}$ + πn = 0 n = positive integer including zero
cos ( $\frac{π t}{2}$ ) = 0 when ( t ) = 1 + 2n
The particle is at rest when ( t ) = 1 + 2n, when n = positive integer including zero.
3c) The particle is at rest when ( t ) = odd positive integer ( 1 , 3 , 5 , … )
d) When is the particle moving in the positive direction?
When is the velocity greater than zero?
3c) The particle is at rest when ( t ) = odd integer
So, the sign changes at every odd integer ( 1, 3, 5, 7, 9,…)
0 ≤ t < 1 = [ 0 , 1 ) = positive direction
1 < t < 3 = ( 1 , 3 ) = negative direction
3 < t < 5 = ( 3 , 5 ) = positive direction
5 < t < 7 = ( 5 , 7 ) = negative direction
7 < t < 9 = ( 7 , 9 ) = positive direction, etc.
3d) [ 0 , 1 ) ∪ ( 3 , 5 ) ∪ ( 7 , 9 ) …
e) Find the total distance traveled in the first 6 seconds.
f ( t ) = sin ( $\frac{π t}{2}$ )
[ 0 , 1 )
f ( 0 ) = sin ( $\frac{π ( 0 ) }{2}$ )
f ( 0 ) = sin ( 0 )
f ( 0 ) = 0
f ( 1 ) = sin ( $\frac{π ( 1 ) }{2}$ )
f ( 1 ) = sin ( $\frac{π}{2}$ )
f ( 1 ) = 1
| f ( 1 ) - f ( 0 ) | =
| 1 - 0 | = 1 foot
Distance traveled in the first second is 1 foot.
( 1 , 3 )
f ( 3 ) = sin ( $\frac{π ( 3 )}{2}$ )
f ( 3 ) = sin ( $\frac{3π}{2}$ )
f ( 3 ) = - 1
| f ( 3 ) - f ( 1 ) | =
| - 1 - 1 | = 2 feet
Distance traveled from 1 second to 3 seconds is 2 feet.
f ( 5 ) = sin ( $\frac{π ( 5 )}{2}$ )
f ( 5 ) = sin ( $\frac{5π}{2}$ )
f ( 5 ) = 1
| f ( 5 ) - f ( 3 ) | =
| 1 – ( - 1 ) | = 2 feet
Distance traveled from 3 seconds to 5 seconds is 2 feet.
f ( 6 ) = sin ( $\frac{π ( 6 )}{2}$ )
f ( 6 ) = sin ( $\frac{6π}{2}$ )
f ( 6 ) = sin ( 3π )
f ( 6 ) = 0
| f ( 6 ) - f ( 5 ) | =
| 0 – 1 | = 1 foot
Distance traveled from 5 seconds to 6 seconds is 1 foot.
[ 0 , 1 ) = 1 foot
( 1 , 3 ) = 2 feet
( 3 , 5 ) = 2 feet
( 5 , 6 ] = 1 foot
[ 0 , 6 ] = 6 feet
3e) The distance traveled in the first 6 seconds is 6 feet.
3f) See sketch of 3f)
g) Find the acceleration at time ( t ) and after 1 second.
v ( t ) = $\frac{π}{2}$ cos ( (π t)/2 )
v ‘ ( t ) = a ( t ) = d/dx [ $\frac{π}{2}$ cos ( $\frac{π t}{2}$ ) ] * d/dx [ $\frac{π}{2}$ t ]
a ( t ) = - $\frac{π}{2}$ sin ( $\frac{π t}{2}$ )( $\frac{π}{2}$ )
3g) a ( t ) = - $\frac{π²}{4}$ sin ( $\frac{π t}{2}$ )
a ( 1 ) = - $\frac{π²}{4}$ sin ( $\frac{π ( 1 ) }{2}$ )
a ( 1 ) = - $\frac{π²}{4}$ sin ( $\frac{π}{2}$ )
a ( 1 ) = - $\frac{π²}{4}$ ( 1 )
3g) a ( 1 ) = - $\frac{π²}{4}$
3h) See photo of 3f)
i) When is the particle speeding up?
The particle is speeding up when v ( t ) and a ( t ) have the same sign.
[ 0 , 1 ) Velocity is positive, acceleration is negative. The signs are different.
( 1 , 2 ) Velocity is negative, acceleration is negative. The signs are the same.
( 2 , 3 ) Velocity is negative, acceleration is positive. The signs are different.
( 3 , 4 ) Velocity is positive, acceleration is positive. The signs are the same.
( 4 , 5 ) Velocity is positive, acceleration is negative. The signs are different.
( 5 , 6 ] Velocity is negative, acceleration is negative. The signs are the same.
3i) The particle is speeding up on the interval ( 1 , 2 ) ∪ ( 3 , 4 ) ∪ ( 5 , 6 ].
3i) The particle is slowing down on the interval [ 0 , 1 ) ∪ ( 2 , 3 ) ∪ ( 4 , 5 ).
