Question

At launch a rocket ship weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 161 km/h; at the end of the first 1.00 min, its speed is 1610 km/h. (a) What is the average acceleration (in m/s2) of the rocket (i) during the first 8.00 s and (ii) between 8.00 s and the end of the first 1.00 min? (b) Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals), what distance does the rocket travel (i) during the first 8.00 s and (ii) during the interval from 8.00 s to 1.00 min?

Answer 1

Answer:

$5.590278\ m/s^2$

$7.74038461538\ m/s^2$

178.888896 m

12790.56 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{161}{3.6}-0}{8}\\\Rightarrow a=5.590278\ m/s^2$

The acceleration is $5.590278\ m/s^2$

$v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{1610}{3.6}-\dfrac{161}{3.6}}{60-8}\\\Rightarrow a=7.74038461538\ m/s^2$

The acceleration is $7.74038461538\ m/s^2$

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 5.590278\times 8^2\\\Rightarrow s=178.888896\ m$

Distance traveled in the first 8 seconds is 178.888896 m

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=\dfrac{161}{3.6}\times 52+\dfrac{1}{2}\times 7.74038461538\times 52^2\\\Rightarrow s=12790.56\ m$

Distance traveled during 8-60 second interval is 12790.56 m