At ( t = 0 ), ( s = 0 \Rightarrow C_2 = 0 ). Thus: [ \boxeds(t) = t^3 ]
[ \int ds = \int 3t^2 , dt ] [ s = t^3 + C_2 ]
Since the particle moves to increasing ( s ) from rest at ( s=1 ), take positive root.
[ \int dv = \int 6t , dt ] [ v = 3t^2 + C_1 ]
Use ( a = v \fracdvds = -0.5v ). Cancel ( v ) (assuming ( v \neq 0 )):
Ground: ( s = 0 ). Use ( v^2 = v_0^2 + 2a(s - s_0) ): [ v^2 = 20^2 + 2(-9.81)(0 - 50) ] [ v^2 = 400 + 981 = 1381 ] [ v = -\sqrt1381 \quad (\textnegative because downward) ] [ \boxedv \approx -37.16 , \textm/s ]
[ v = v_0 + at ] [ s = s_0 + v_0 t + \frac12 a t^2 ] [ v^2 = v_0^2 + 2a(s - s_0) ]
At ( t = 0 ), ( s = 0 \Rightarrow C_2 = 0 ). Thus: [ \boxeds(t) = t^3 ]
[ \int ds = \int 3t^2 , dt ] [ s = t^3 + C_2 ] rectilinear motion problems and solutions mathalino
Since the particle moves to increasing ( s ) from rest at ( s=1 ), take positive root. At ( t = 0 ), ( s = 0 \Rightarrow C_2 = 0 )
[ \int dv = \int 6t , dt ] [ v = 3t^2 + C_1 ] Cancel ( v ) (assuming ( v \neq 0 )): Ground: ( s = 0 )
Use ( a = v \fracdvds = -0.5v ). Cancel ( v ) (assuming ( v \neq 0 )):
Ground: ( s = 0 ). Use ( v^2 = v_0^2 + 2a(s - s_0) ): [ v^2 = 20^2 + 2(-9.81)(0 - 50) ] [ v^2 = 400 + 981 = 1381 ] [ v = -\sqrt1381 \quad (\textnegative because downward) ] [ \boxedv \approx -37.16 , \textm/s ]
[ v = v_0 + at ] [ s = s_0 + v_0 t + \frac12 a t^2 ] [ v^2 = v_0^2 + 2a(s - s_0) ]