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

Related Maths A Level answers

All answers ▸

The curve C has the equation y=((x^2+4)(x-3))/2*x where x is not equal to 0 . Find the tangent to the curve C at the point where x=-1 in the form y=mx+c


Find the integral of (3x^2+4x^5-7)dx


What is the smallest possible value of the integral ∫(x-a)^2 dx between 0 and 1 as a varies?


Given that y = 5x^2 - 4/(x^3), x not equal to 0, find dy/dx.