A car traveling 60 mph (88 ft/sec) undergoes a constant deceleration until it comes to rest approximately 9.09 sec later. The distance d ( t ) ( in ft ) that the car travels t seconds after the brakes are applied is given by d ( t ) = − 4.84 t 2 + 88 t , where 0 ≤ t ≤ 9.09 . ( See Example 5 ) a. Find the difference quotient d ( t + h ) − d ( t ) h . Use the difference quotient to determine the average rate of speed on the following intervals for t . b. [ 0 , 2 ] ( Hint : t = 0 and h = 2 ) c. [ 2 , 4 ] ( Hint : t = 2 and h = 2 ) d. [ 4 , 6 ] ( Hint : t = 4 and h = 2 ) e. [ 6 , 8 ] ( Hint : t = 6 and h = 2 )
A car traveling 60 mph (88 ft/sec) undergoes a constant deceleration until it comes to rest approximately 9.09 sec later. The distance d ( t ) ( in ft ) that the car travels t seconds after the brakes are applied is given by d ( t ) = − 4.84 t 2 + 88 t , where 0 ≤ t ≤ 9.09 . ( See Example 5 ) a. Find the difference quotient d ( t + h ) − d ( t ) h . Use the difference quotient to determine the average rate of speed on the following intervals for t . b. [ 0 , 2 ] ( Hint : t = 0 and h = 2 ) c. [ 2 , 4 ] ( Hint : t = 2 and h = 2 ) d. [ 4 , 6 ] ( Hint : t = 4 and h = 2 ) e. [ 6 , 8 ] ( Hint : t = 6 and h = 2 )
Solution Summary: The author calculates the difference quotient of the distance function d(t)=-4.84t
A car traveling 60 mph (88 ft/sec) undergoes a constant deceleration until it comes to rest approximately 9.09 sec later. The distance
d
(
t
)
(
in ft
)
that the car travels t seconds after the brakes are applied is given by
d
(
t
)
=
−
4.84
t
2
+
88
t
, where
0
≤
t
≤
9.09
. (See Example 5 )
a. Find the difference quotient
d
(
t
+
h
)
−
d
(
t
)
h
.
Use the difference quotient to determine the average rate of speed on the following intervals for t.
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