Solve the problem.
-An arrow is shot straight up from level ground. The distance (in meters) of the arrow above the ground (the position function) is $\mathrm { f } ( \mathrm { t } ) = 4 + 130 \mathrm { t } - 4.9 \mathrm { t } ^ { 2 }$ at any time $\mathrm { t }$ (in sec). Find $\mathrm { f } ^ { \prime } ( 1 )$ and the initial velocity of the arrow.
A) $\mathrm { f } ^ { \prime } ( 1 ) = 120.2$; initial velocity $= 130 \mathrm {~m} / \mathrm { s }$
B) $\mathrm { f } ^ { \prime } ( 1 ) = 124.2$; initial velocity $= 130 \mathrm {~m} / \mathrm { s }$
C) $f ^ { \prime } ( 1 ) = 120.2$; initial velocity $= 4 \mathrm {~m} / \mathrm { s }$
D) $f ^ { \prime } ( 1 ) = - 9.8$; initial velocity $= 130 \mathrm {~m} / \mathrm { s }$

Question

Quizplus · Accepted Answer

The derivative of the position function $f(t) = 4 + 130t - 4.9t^2$ with respect to time $t$ gives the velocity function $f'(t) = 130 - 9.8t$. Evaluating $f'(1) = 130 - 9.8(1) = 120.2$ m/s. The initial velocity is the coefficient of $t$ in the original equation, which is $130$ m/s.

Solve the Problem f(t)=4+130t−4.9t2\mathrm { f } ( \mathrm { t } ) = 4 + 130 \mathrm { t } - 4.9 \mathrm { t } ^ { 2 }f(t)=4+130t−4.9t2

Solve the Problem $\mathrm { f } ( \mathrm { t } ) = 4 + 130 \mathrm { t } - 4.9 \mathrm { t } ^ { 2 }$