Question

A particle is subjected to an acceleration a= alpha *t +beta*t², where alpha and beta are constants. the position and vel.ocity of the particle at t=0 are x₀ and v₀ respectively. what are the expressions for position x and velocity v of the particle at time t?

Answer 1

Dear student

$$\begin{gathered} Giventheaccelerationis \\ a=\alpha t+\beta t^2 \\ butaccelerationisgivenby\frac{d^{2}x}{d{t}^2}idifferentialformso \\ \Rightarrow \frac{d^{2}x}{d{t}^2}=\alpha t+\beta t^2 \\ integrateaboveequatiomw.r.t.t \\ \Rightarrow \frac{dx}{dt}={\frac{\alpha t}{2}}^2+\frac{\beta t^3}{3}+c \\ wherecisthecons\tan tofintegral.\frac{dx}{dt}isthevelocityoftheparticlev.so \\ \Rightarrow v={\frac{\alpha t}{2}}^2+\frac{\beta t^3}{3}+c \\ butatt=0,v=v_{o}so \\ \Rightarrow v_o=0+0+c \\ \Rightarrow v_o=c \\ Hence \\ \Rightarrow v={\frac{\alpha t}{2}}^2+\frac{\beta t^3}{3}+v_o \\ Integratethe\frac{dx}{dt}again \\ \Rightarrow x={\frac{\alpha t}{6}}^3+\frac{\beta t^4}{12}+v_{o}t+c\text{'} \\ wherec\text{'}isthecons\tan tofintegral.Butatt=0,x=x_{o}so \\ \Rightarrow x_o=0+0+0+c\text{'} \\ \Rightarrow x_o=c\text{'} \\ \Rightarrow x={\frac{\alpha t}{6}}^3+\frac{\beta t^4}{12}+v_{o}t+x_o \end{gathered}$$