At ( t = 0 ), ( s = 0 \Rightarrow C_2 = 0 ). Thus: [ \boxeds(t) = t^3 ]
[ \fracdvv = -0.5 , dt ] Integrate: [ \ln v = -0.5t + C ] At ( t=0, v=20 \Rightarrow \ln 20 = C ). [ \ln\left( \fracv20 \right) = -0.5t ] [ \boxedv(t) = 20e^-0.5t ]
Use ( a = v \fracdvds = -0.5v ). Cancel ( v ) (assuming ( v \neq 0 )): rectilinear motion problems and solutions mathalino
Use ( v = v_0 + at ): [ 0 = 20 - 9.81 t \quad \Rightarrow \quad t = \frac209.81 \approx \boxed2.038 , \texts ]
[ \int ds = \int 3t^2 , dt ] [ s = t^3 + C_2 ] At ( t = 0 ), ( s = 0 \Rightarrow C_2 = 0 )
Since the particle moves to increasing ( s ) from rest at ( s=1 ), take positive root.
We know ( v = \fracdsdt = 3t^2 ). Integrate: Cancel ( v ) (assuming ( v \neq
Topics: Dynamics, Engineering Mechanics, Calculus-Based Kinematics What is Rectilinear Motion? Rectilinear motion refers to the movement of a particle along a straight line. In engineering mechanics, this is the simplest form of motion. The position of the particle is described by its coordinate ( s ) (often measured in meters or feet) along the line from a fixed origin.