Answer
4a) v ( t ) = $\frac{2t - t²}{e^t}$
4b) v ( 1 ) = $e^{- 1}$
4c) The particle is at rest at zero seconds and 2 seconds.
4d) The particle is moving in the positive direction on the interval 0 < t < 2 .
4e) The distance traveled in the first 6 seconds is ≈ 0.933 feet.
4f) See sketch of 4f)
4g) a ( t ) = $\frac{t² - 3t + 2}{e^t}$
4g) a ( 1 ) = 0
4h) See photo of 4h)
4i) The particle is speeding up on the interval ( 0 , 1 ).
4i) The particle is slowing down on the interval ( 1 , 6 ).
Work Step by Step
4) f ( t ) = t² $e^{-t}$
a) Find the velocity at time ( t ).
f ( t ) = $\frac{t²}{e^t}$
Quotient Rule
f ‘( t ) = v ( t ) = $\frac{d/dx[ t² ]( e^t) - d/dx [ e^t ]( t² ) }{( e^t )²}$
v ( t ) = $\frac{( 2t )( e^t) - ( e^t )( t² ) }{( e^t )²}$
v ( t ) = $\frac{( e^t)[ ( 2t ) - ( t² ) ]}{( e^t )²}$
4a) v ( t ) = $\frac{2t - t²}{e^t }$
v ( 1 ) = $\frac{2( 1 ) - ( 1 )²}{e^1 }$
v ( 1 ) = $\frac{2 - 1}{e }$
v ( 1 ) = $\frac{1}{e }$ = $e^{-1}$
4b) v ( 1 ) = $e^{-1}$
When is velocity equal to zero?
v ( t ) = $\frac{2t - t²}{e^t }$
0 = $\frac{2t - t²}{e^t }$
The denominator ( $e^{t}$ ) is always greater than zero so ( t ) can be all real numbers. If the numerator ( 2t – t² ) equals zero, then the velocity function equals zero.
0 = 2t – t²
0 = – t² + 2t
( - 1 ) 0 = ( - 1 ) ( – t² + 2t )
0 = t² - 2t
0 + ? = t² - 2t + ?
- 2 * 1/2 = 1
1² = 1
0 + ( 1 ) = t² - 2t + ( 1 )
1 = ( t – 1 )²
$\sqrt 1$ = $\sqrt ( t – 1 )²$
± 1 = t – 1
1 ± 1 = t
2 = t
0 = t
4c) The particle is at rest at zero seconds and 2 seconds.
When is the particle moving in the positive direction?
When is velocity greater than zero?
When ( t ) = 0 , or ( t ) = 2 , the particle changes direction.
When 0 < t < 2 , the numerator ( 2t – t² ) is positive and the denominator ( $e^{t}$ )is positive, so the velocity function is positive.
When t > 2 the numerator ( 2t – t² ) is negative and the denominator ( $e^{t}$ )is positive, so the velocity function is negative.
4d) The particle is moving in the positive direction on the interval 0 < t < 2 .
Find the total distance traveled in the first 6 seconds.
f ( t ) = $\frac{t²}{e^t}$
( 0 , 2 )
f ( 0 ) = $\frac{( 0 )²}{e^( 0 )}$
f ( 0 ) = $\frac{0}{1}$ = 0
f ( 2 ) = $\frac{( 2 )²}{e²}$
f ( 2 ) = $\frac{4}{e²}$
| f ( 2 ) – f ( 0 ) | =
| ($\frac{4}{e²}$ ) - 0 | = $\frac{4}{e²}$ ≈ 0.541
The distance traveled in the first 2 seconds is $\frac{4}{e²}$ ≈ 0.541 feet.
f ( 6 ) = $\frac{6²}{e^6}$
f ( 6 ) = $\frac{36}{e^6}$
| f ( 6 ) – f ( 2 ) | =
| $\frac{36}{e^6}$ – $\frac{4}{e²}$ | ≈ | 0.089 – 0.541 | ≈ 0.452 feet
The distance traveled from 2 seconds to 6 seconds is ≈ 0.452 feet.
( 0 , 2 ) ≈ 0.541 feet
( 2 , 6 ] ≈ 0.452 feet
( 0 , 6 ] ≈ 0.933 feet
4e) The distance traveled in the first 6 seconds is ≈ 0.933 feet.
4f) See sketch of 4f)
Find the acceleration at time ( t ) and after 1 second.
v ( t ) = $\frac{2t - t²}{e^t}$
v ‘( t ) = a ( t ) = $\frac{d/dx [2t - t² ]( e^t ) - d/dx [ e^t ]( 2t - t²)}{( e^t )²}$
a ( t ) = $\frac{(2 - t )( e^t ) - ( e^t )( 2t - t²)}{( e^t )²}$
a ( t ) = $\frac{( e^t ) [(2 - t ) - ( 2t - t²) ]}{( e^t )²}$
a ( t ) = $\frac{(2 - t ) - ( 2t - t²)}{ e^t }$
a ( t ) = $\frac{2 - t - 2t + t²}{ e^t }$
a ( t ) = $\frac{t² - 3t + 2}{ e^t }$
4g) a ( t ) = $\frac{( t - 1 )( t - 2 )}{ e^t }$
a ( 1 ) = $\frac{( ( 1 ) - 1 )( ( 1 ) - 2 )}{ e^1 }$
a ( 1 ) = $\frac{( 0 )( ( - 1 )}{ e}$ = 0
4g) a ( 1 ) = 0
4h) See photo for 4h)
When is the particle speeding up? When is it slowing down?
When the velocity and the acceleration have the same sign, the particle is speeding up.
a ( t ) = $\frac{( t - 1 )( t - 2 )}{ e^t }$
0 = $\frac{( t - 1 )( t - 2 )}{ e^t }$
The denominator ( e^t ) is positive for every t-value.
0 = ( t – 1 )( t – 2 )
t = 1
t = 2
a ( t ) is equal to zero at 1 second and at 2 seconds.
( t – 1 )( t – 2 ) > 0
When 0 < t < 1 , ( t – 1 ) is negative and ( t – 2 ) is negative so ( t – 1 )( t – 2 ) is positive.
When 1 < t < 2 , ( t – 1 ) is positive and ( t – 2 ) is negative so ( t – 1 )( t – 2 ) is negative.
When t > 2 , ( t – 1 ) is positive and ( t – 2 ) is positive so ( t – 1 )( t – 2 ) is positive.
Acceleration is positive ( 0 , 1 ) ∪ ( 2 , ∞ )
Velocity is positive ( 0 , 2 )
( 0 , 1 ) Velocity is positive and acceleration is positive, the signs are the same.
( 1 , 2 ) Velocity is positive and acceleration is negative, the signs are different.
( 2 , 6 ) Velocity is negative and acceleration is positive, the signs are different.
4i) The particle is speeding up on the interval ( 0 , 1 ).
4i) The particle is slowing down on the interval ( 1 , 6 ).
