Q. 69 - Calculus

Question

The initial velocity of the ball is 42 feet per second, released from a height of 4 feet.

The acceleration due to gravity is a(t)= -32 feet per second square.

(a) Calculate ${\int }_{}^{}a\left(t\right)$  and then to find a formula for the velocity v(t) of the ball after t seconds.

(b)  Calculate  $\int v\left(t\right)dt$then to find a formula for the velocity of the ball seconds and its initial position was s ₀= 4 feet.

Step-by-Step Solution

Answer

Part(a) (i) $-32t$

(ii)$v \left(t\right)=-32t+42$

Part (b) (i)$-16t^2+42t$

(ii)$s\left(t\right)=-16t^2+42t+4$

Step 1. Given

The initial velocity of the ball is 42 feet per second, released from a height of 4 feet.

The acceleration due to gravity is a(t)= -32 feet per second square.

Part(a) Step 2. Calculation of ∫ a ( t ) .

$$\begin{gathered} {\int }_{}^{}a\left(t\right)dt \\ ={\int }_{}^{}-32dt \\ =-32{\int }_{}^{}dt \\ =-32t \end{gathered}$$

Step 3. Calculation of the formula of velocity

$$\begin{gathered} \mathrm{Since} , \mathrm{acceleration} \mathrm{is} \mathrm{the} \mathrm{derivative} \mathrm{of} \mathrm{velocity} \mathrm{so} \mathrm{it} \mathrm{means} \mathrm{that} \mathrm{velocity} \mathrm{is} \mathrm{the} \mathrm{integration} \mathrm{of} \mathrm{acceleration} \\ v \left(t\right) ={\int }_{}^{}a\left(t\right)dt + v_0 \\ v\left(t\right)=-32t+42 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{formula} \mathrm{for} \mathrm{velocity} \mathrm{i}s v \left(t\right)=-32t+42 \end{gathered}$$

Part(b) Step 4. Calculation of ∫ v ( t ) d t and formula for s(t)

$$\begin{gathered} \int v\left(t\right)dt \\ \int -32t+ 42dt \\ -32\int tdt+42\int dt \\ -16t^2+42t \\ s\left(t\right)=-16t^2+42t+4 \end{gathered}$$

Hence the formula for $s\left(t\right)=-16t^2+42t+4$