Question

What is the magnitude of the acceleration that the engine must produce

Answer 1

ANSWER

$\begin{equation*} 111.6\text{ }m\/s^2 \end{equation*}$

EXPLANATION

To find the magnitude of the acceleration, find the components of the acceleration on the x and y-axis.

In the x-direction, we can find the acceleration using the formula:

$x=x_0+v_0t+\frac{1}{2}a_xt^2$

where t = time taken

x = distance traveled in the x-direction

a = acceleration in the x-direction

v0 = initial velocity in the x-direction

x0 = 0 m

Hence, the acceleration in the x-direction is:

$\begin{gathered} 19500\cos32=0+(1810\cos20)(9.20)+\frac{1}{2}*a_x*(9.20)^2 \\ 16536.94=15647.76+42.32a_x \\ 42.32a_x=16536.94-15647.76=889.18 \\ a_x=\frac{889.18}{42.32} \\ a_x=21.01\text{ }m\/s^2 \end{gathered}$

In the y-direction, we can find the acceleration using the formula:

$y=y_0+v_0t+\frac{1}{2}a_yt^2$

where y = distance traveled in the y-direction

y0 = 0 m

Hence, the acceleration in the y-direction is:

$\begin{gathered} 19500\sin32=0+(1810\sin20)(9.20)+\frac{1}{2}a_y*9.20^2 \\ 10333.43=5695.32+42.32a_y \\ 42.32a_y=10333.43-5695.32=4638.11 \\ a_y=\frac{4638.11}{42.32} \\ a_y=109.6\text{ }m\/s^2 \end{gathered}$

The magnitude of the acceleration is given by:

$a=\sqrt{a_x^2+a_y^2}$

Therefore, the magnitude of the acceleration is:

$\begin{gathered} a=\sqrt{21.01^2+109.6^2} \\ a=\sqrt{441.4201+12012.16}=\sqrt{12453.5801} \\ a=111.6\text{ }m\/s^2 \end{gathered}$

That is the answer.