A particle is moving in a straight line from A to B with constant acceleration 4m/s^2. The velocity of the particle at A is 3m/s in the direction AB. The velocity of the particle at B is 18m/s in the same direction/ Find the distance from A to B.

First draw a diagram to see the set-up.Then look at SUVAT to see which values we have been given. In this case it is a=4, u=3,v=18 and s=?. The only letter not used from SUVAT is the t so we use the formula without... v2=u2+2as. Fill in the numbers 182=32+2 x 4 x s324 = 9+ 8s. Rearranges = (324-9)/8 = 39.375 m

AK
Answered by Adam K. Further Mathematics tutor

2878 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

x^3 + 2x^2 - 9x - 18 = (x^2 - a^2)(x + b) where a,b are integers. Work out the three linear factors of x^3 + 2x^2 - 9x - 18. (Note: x^3 indicates x cubed and x^2 indicates x squared).


If y=x^3+9x, find gradient of the tangent at (2,1).


Using differentiation, show that f(x) = 2x^3 - 12x^2 + 25x - 11 is an increasing function.


If y=(x^2)*(x-10), work out dy/dx


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:

© 2025 by IXL Learning