Question

The last frame before the shuttle begins to move is 140. The shuttle has traveled 56 meters in 243 frames.(a) If each frame is 24 seconds what is the time elapsed? =4.29 seconds (b) Assuming constant acceleration, at what rate is the shuttle accelerating?(c) If the shuttle continued to move with this acceleration, what speed would it reach 76s after launch?(d) If the shuttle traveled directly upwards,what would its altitude be at 76 s?

Answer 1

ANSWER:

(b) 6.09 m/s²

(c) 462.84 m/s

(d) 28302.4 m

STEP-BY-STEP EXPLANATION:

Given:

Time elapsed (t) = 4.29

Distance (d) = 56 m

initial velocity (u)= 0 m/s

(b)

We can determine the acceleration by the following formula:

$\begin{gathered} s=ut+\frac{1}{2}at^2 \\ \\ \text{ we replacing} \\ \\ 56=0\cdot4.29+\frac{1}{2}(a)(4.29)^2 \\ \\ \frac{18.4041}{2}a=56 \\ \\ a=\frac{56\cdot2}{18.4041} \\ \\ a=6.09\text{ m/s}^2 \end{gathered}$

(c)

Now we calculate the speed as follows:

$\begin{gathered} v=u+at \\ \\ \text{ We replacing} \\ \\ v=0+6.09(76) \\ \\ v=462.84\text{ m/s} \end{gathered}$

(d)

We can calculate the altitude using the following formula:

$\begin{gathered} h=ut+\frac{1}{2}gt^2 \\ \\ \text{ We replacing} \\ \\ h=0\cdot76+\frac{1}{2}(9.8)(76)^2 \\ \\ h=28302.4\text{ m} \end{gathered}$