Question

A car which is traveling at a velocity of 27 m/s undergoes an acceleration of 5.5 m/s^2 over a distance of 430 m. How fast is it going after that acceleration?

Answer 1

73.9m/s (1dp)

1) list everything that you are given using suvat, where s is distance, u is initial velocity(speed), v is final velocity(speed), a is acceleration and t is time

s = 430m

u = 27 m/s

v = ? (we need to work out)

a = 5.5m/s^2

t = (we are not given this value)

2) use an equation that doesn't involve the time

${v}^{2} = u {}^{2} + 2as$

3) input the values that we have

${v}^{2} = ( {27})^{2} + 2(5.5)(430)$

$v {}^{2} = 5459$

$v = \sqrt{5459} = 73.9$

so the answer is 73.9m/s to 1dp