$\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}$

Differentiate both sides of the obtained equation with respect to t.

$\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}$

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.

$\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}$

The initial position is obtained at t=0. Substitute t=0 in the given position function.

$\begin{gathered} s(0)=-23\times0+65 \\ =65 \end{gathered}$

8 0

1 year ago

4 bells toll together at 9 am. They toll after 7, 8. 11 and 12 seconds respectively. How many times will they toll together agai

Svetach [21]

Answer:

5 times

Step-by-step explanation:

What we need to do here is to find the lowest common multiple of 7,8,11 and 12

This is;

1848

Now, let’s calculate the number of seconds in 3 hours

In 1 hour, we have 3600 seconds

In 3 hours, this will be;

3600 * 3 = 10,800 seconds

So, to find the number of times in which all 4 will toll together, we need to know the number of 1848 seconds in 10,800 seconds

Mathematically, that will be;

10,800/1848 = 5.844

Our answer will be 5 times however

This is because the sixth time they would have sounded together would be more than the 3 hours time frame

8 0

3 years ago