Question

At launch, a rocket ship weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 ss to reach 155 km/h; at the end of the first 1.00 min, its speed is 1550 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

A. I. 5.38m/s

II. 8.28 m/s

B. I. 344.5 m or 0.3445 km

II. 22391.2 m or 22.391 km

Step-by-step explanation:

A. I. First 8.00 secs

Speed in m/s = (155 * 1000) / 3600 = 43.06 m/s

Acceleration = speed/time = 43.06/8 = 5.38 m/s²

II. Between 8 secs and 1 min, time taken is 52 secs

Speed = (1550 * 1000) / 3600 = 430.6 m/s

Acceleration = speed/time = 430.6/52 = 8.28 m/s²

B. I. During the first 8 secs, assuming constant acceleration:

Distance = speed * time = 43.06 * 8 = 344.48 m = 0.3445 km

II. Between 8secs and 1 min, assuming constant acceleration:

Distance = speed * time = 430.6 * 52 = 22391.2m = 22.391 km