A rocket travels at 500m/s two minutes after its take-off. If it was initially stationary, calculate its acceleration. If the rocket has a mass of 1800kg, what force is required to give it an acceleration of 2m/s^2?

a) Acceleration is the rate of change of velocity of an object

In this case, the equation that we will use to calculate it is 

a = (v - u )/ t

where v is the rocket's final speed, u is its initial speed, and t is the time taken

Note: units of measurement 

a: measured in m/s^2

v/u: measured in m/s

t: measured in s

Putting the given values into this equation, we find that 

acceleration = (500 - 0)m/s / (2x60)s = 4.17 m/s^2 (2dp)

- Hints: the initial speed of the rocket was 0 m/s at it was stationary to begin with, and we multiply 2 by sixty as we must convert time values from minutes into seconds. 

- I have included units in the calculation above- you dont need to do this at every stage of calculation but make sure you include units when giving a final answer!

b)  We can work out the second part of this question using Newtons's second law of motion F = m x a 

 F = 1800 x 2   = 3600 N ( or 3600 kgm/s^2) 

This law explains that the force (F) acting on an object is equal to the mass (m) of an object times its acceleration (a)

DS
Answered by Dhrushee S. Physics tutor

5236 Views

See similar Physics GCSE tutors

Related Physics GCSE answers

All answers ▸

What provides the centripetal force on a satellite and what are the factors that determine the size of the centripetal force on the satellite


How do current and voltage vary in series and parallel circuits, respectively?


Can you please explain the basics of electricity? I can do the maths but I don't understand what 'voltage', 'current' or 'resistance' actually is!


A student investigated how the resistance of a piece of nichrome wire varies box with length.Describe how the student would obtain the data needed for the investigation. Your answer should include a risk assessment for one hazard in the investigation.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning