Solve the problem.
-Given the acceleration, initial velocity, and initial position of a body moving along a coordinate line at time t, find the body's position at time t. $$a = 32 , v ( 0 ) = 13 , s ( 0 ) = - 2$$
A) $s = 32 t ^ { 2 } + 13 t - 2$
B) $s = - 16 t ^ { 2 } - 13 t - 2$
C) $\mathrm { s } = 16 \mathrm { t } ^ { 2 } + 13 \mathrm { t } - 2$
D) $\mathrm { s } = 16 \mathrm { t } ^ { 2 } + 13 \mathrm { t }$

Question

Quizplus · Accepted Answer

We can use the formulas of motion to find the position function. Integrating the acceleration function gives us the velocity function:
$$v(t) = \int a dt = 32t + C_1$$
To find the constant $C_1$, we use the initial velocity condition:
$$v(0) = 32(0) + C_1 = 13 \implies C_1 = 13$$
So, we have $v(t) = 32t + 13$. Integrating the velocity function gives us the position function:
$$s(t) = \int v(t) dt = \int (32t + 13) dt = 16t^2 + 13t + C_2$$
To find the constant $C_2$, we use the initial position condition:
$$s(0) = 16(0)^2 + 13(0) + C_2 = -2 \implies C_2 = -2$$
So, we have $s(t) = 16t^2 + 13t - 2$. Therefore, the best choice is (C).

Solve the Problem a=32,v(0)=13,s(0)=−2a = 32 , v ( 0 ) = 13 , s ( 0 ) = - 2a=32,v(0)=13,s(0)=−2

Solve the Problem $a = 32 , v ( 0 ) = 13 , s ( 0 ) = - 2$