Constant acceleration (suvat) equationsAQA A-Level Maths: Flashcards

What these 13 flashcards ask

State the five constant acceleration equations.
When can the suvat equations be used?
Which equation has no s?
Which equation has no t?
Which equation has no v?
Which equation has no a?
What does "starts from rest" mean for suvat?
What does "comes to rest" mean for suvat?
How do you derive v=u+at?
How do you derive s=ut+\frac12at^2?
How do you derive v^2=u^2+2as?
What does a negative a mean?
When does one object overtake another?

Exam questions on Constant acceleration (suvat) equations

A car travels in a straight line with constant acceleration. Over a $6$ s period its speed increases from $8$ m s $^{-1}$ to $20$ m s $^{-1}$ .
Find the speed of the car when it has travelled $50$ m from the start of the $6$ s period.2 marks
A train slows down with constant deceleration along a straight track. It passes a signal at $30$ m s $^{-1}$ and comes to rest $450$ m beyond the signal.
Find the speed of the train when it is $200$ m beyond the signal.2 marks
A particle moves in a straight line with constant acceleration $a$ . At time $t=0$ its velocity is $u$ . At time $t$ its velocity is $v$ and its displacement from its starting point is $s$ .
Using the definition of acceleration and the fact that the average velocity is $\frac12(u+v)$ for constant acceleration, derive the formula $s=ut+\frac12at^2$ .3 marks

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).