A car is accelerating at 2 ms^-2 along a horizontal road. It passes a point A with a velocity of 10 ms^-1 and later a point B, where AB = 50m. FInd the velocity of the car as it passes through B.

From the first look, we can see the question states the car is undergoing constant acceleration, thus we know the SUVAT equations are valid. Then reading on, we're required to find the final velocity, v, of the car as it passes B, therefore we must find at least 3 of the 4 other variables of the system (a, u, t and S in addition to v). As we are given a = 2ms^-2, u = 10ms^-1 (the velocity at A) and S = 50m, we can put these values straight into any SUVAT equation containing a, u, v and S. This equation is v^2 = u^2 +2as. Therefore the answer is:
v = sqrt(u^2 + 2as)v = sqrt(10^2 + 2250)v = 10*sqrt(3) = 17.32 ms^-1 to 2 d.p

Answered by Maths tutor

3020 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find, using calculus, the x coordinate of the turning point of the curve y=e^(3x)*cos(4x) pi/4<x<pi/2 (Edexcel C3)


AQA PC4 2015 Q5 // A) Find the gradient at P. B) Find the equation of the normal to the curve at P C)The normal P intersects at the curve again at the point Q(cos2q, sin q) Hence find the x-coordinate of Q.


Do y=3x^2+5x+12 and y=3x-8 intercept with each other? If yes, at which point(s)?


At each point P of a curve for which x > 0 the tangent cuts the y-axis at T, and N is the foot of the perpendicular from P to the y-axis. If T is always 1 unit below N and the curve passes through the point (1,0), find the Cartesian equation of the curve.


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