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 ▸How to integrate ln(x)
Given a fixed parabola and a family of parallel lines with given fixed gradient, find the one line that intersects the parabola in one single point
(4-2x)/(2x+1)(x+1)(x+3) = A/(2x+1)+B/(x+1)+C(x+3) Find the values of the constants A, B and C
Let f(x) = x * sin(2x). Find the area beneath the graph of y = f(x), bounded by the x-axis, the y-axis and the line x = π/2.