The distance traveled by a passenger train is given by the equation x = 3t^2 + 2t + 4 where x is the distance in kilometer and t is the time in hours. What is the velocity after traveling 10 kilometers?
Velocity is equal to first derivative of distance with respect to time, so,
v = dx / dt
x = 3t^2 + 2t + 4 (eq.1)
dx/dt = v = 6t + 2
v = 6t + 2 (eq.2)
find t to solve v
from the eq.1 solve the value of t where x = 10 km
10 = 3t^2 + 2t + 4
3t^2 + 2t – 6 = 0
solve t using Quadratic Formula
t = [-b ± (b^2 - 4ac)^0.5] / 2a
where a, b and c are the coefficient of t^2, t and constant term.
t = 1.11963 hrs and −1.7863 hrs
used positive value of t to solve the value of velocity after 10 km.
from eq.2 solve the value of v when t = 1.11963 hrs
v = 6(1.11963) + 2
v = 8.72kph