Question

The acceleration of a particle is defined by the acceleration of a particle is defined by the relation a= kt^2.(a) knowing that v= -10 m/s when t= 0 and that v=10m/s when t=2s.(b) determine the constant k. write the equations of motion, knowing also that x=0 when t=2s.

Answer 1

Question34042

a) We are given initial conditions $v|_{t=0} = -10\frac{m}{s}; v|_{t=2} = 10\frac{m}{s}$ . Acceleration by definition is $a = \frac{dv}{dt}$ , hence integrating this equation gives $v(t) = \int_0^t a(t') dt' = k \int_0^t t'^2 dt' = \frac{kt^3}{3} + C$ . Plugging in initial conditions gives $-10 = C; 10 = \frac{8}{3} k + C$ , which yields $C = -10; k = \frac{30}{4}$ . Thus, velocity as a function of time is $v(t) = \frac{30}{4} \frac{t^3}{3} - 10 = 2.5 t^3 - 10$ .

b) Velocity is by definition $v(t) = \frac{dx}{dt}$ . Knowing law of velocity from previous task, and integrating previous expression, obtain $x(t) = \int_0^t v(t') dt' = \frac{10}{4} \frac{t^4}{4} - 10t + C' = \frac{5}{8} t^4 - 10t + C'$ . From initial condition $x|_{t=2} = 0$ derive $0 = \frac{5}{8} \cdot 16 - 10 \cdot 2 + C' \Rightarrow C' = 10$ . Thus, finally equation of motion is $x(t) = \frac{5}{8} t^4 - 10t + 10$ .