The acceleration of a particle is defined by the relation a=kt^2.?
a = k*t^2 v = int(a dt) v =1/3*k*t^3 + C1 when t = 0, v = v0. v0 = C1 therefore: v = 1/3*k*t^3 + v0 when t=t1, v=v1: v1 = 1/3*k*t1^3 + v0 solve for k: k = 3*(v1-v0)/t1^3 Re-construct our velocity equation: v = (v1-v0)*(t/t1)^3 + v0 integrate to find x: x = int(v, t) x = 1/4*(v1-v0)*t^4/t1^3+v0*t + C2 using x=0 at t = t1: 0 = 1/4*(v1-v0)*t1^4/t1^3+v0*t1 + C2 Solve for C2: C2 = -1/4*t1*v1-3/4*v0*t1 Re-construct x: x = 1/4*(v1-v0)*t^4/t1^3+v0*t -1/4*t1*v1 - 3/4*v0*t1 Data: v0:=-32 ft/s; v1:=32 ft/s; t1:=4 s; Results: (A) k = 3 ft/s^4 (B) Here are all kinematics formulas for this situation a(t) = (3 ft/s^4)*t^2 v(t) = (1ft/s^4) t^3 - 32 ft/sec x(t) = (1/4 ft/s^4)*t^4 - (32 ft/sec)*t + 64 ft
Suppose that the equation of motion for a particle?
Suppose that the equation of motion for a particle (where s is in meters and t in seconds) is s= t^4-2t^3+8 a. Find the time where acceleration is zero (besides at t=0) b.Find the position and velocity at this time
The acceleration of a particle is defined by the relation a= k t^2.?
a=kt² ; so v = ∫a.dt = ⅓.k.t³ + C₁and s = ∫v.dt = k.t⁴/12 + C₁.t + C₂ ; Given s=24 at t=0 we get: C₂= 24 1. From the condition for s given in 1 we have: 96 = k.6⁴/12 + 6.C₁+ 24 or k.108 + 6.C₁= 72 .... (1) From the condition for v given in 1 we have: 18 = 72.k + C₁= 18 giving C₁= 18-72.k Substituting this into (1) gives 108.k + 6.(18-72.k) = 72 or 3k + 3 – 12.k = 2 so that k=1/9 2. Hence C₁= 10 giving v = t³/27 +10 which at t=2 gives v = 10.3 m/s
The equation of motion for a particle is given by s= 2t^3 - 9t^2 - 24t , t >_ 0?
a) Velocity is just the time-derivative of position. s'(t) = 6t^2 - 18t - 24. b) Acceleration is just the time-derivative of velocity (or the second time-derivative of position.) s''(t) = 12t - 18. c) s''t(1) = 12(1) - 18 = -6 meters/sec^2. d) 6t^2 - 18t - 24 = 0 t^2 - 3t - 4 = 0. (t - 4)(t + 1) = 0. Since we're just looking at t > o here, we want the t = 4 solution. At t = 4, the acceleration is s''t(4) = 12(4) - 48 = 0. The acceleration is 0.
A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet.?
f(t) = t^3 − 6t^2 + 9t (a) velocity @ time, t = v(t) v(t) = f'(t) Where f'(t) = 3*t^(3-1) - 6a2t^(2-1) + 9*1t^(1-1) = 3t^2 - 12t^1 + 9t^0 [where a^0 = 1], = 3t^2 - 12t + 9 = 3[t^2 - 4t + 3] v(t) = 3[t^2 - 4t + 3] b) v(2) = f'(2) = 3 [ (2)^2 - 4(2) + 3 ] = 3 [ 4 - 8 + 3 ] = 3 [ -4 + 3 ] = 3 [ -1 ] = -3 v(2) = -3 ft/s [ Recall that velocity is rate of change of displacement with time and has magnitude, (3ft/s) and direction (-ve sign) ] Hence, v(2) = 3 ft/s in the negative direction. c) @ rest, v = 0 ft/s => f'(t) = 0 3 [ t^2 - 4t + 3 ] = 0 Divide both sides by 3, t^2 - 4t + 3 = 0 t^2 - 3t - t + 3 = 0 t(t - 3) - 1(t - 3) = 0 (t - 1)(t - 3) = 0 t - 1 = 0 and t - 3 = 0 t = 1, 3 [ t, v(t) ] = [1, 0], [3, 0] Thus, t = 1s, 3s d) Positive direction = when v(t) > 0 = f'(t) > 0 3 [ t^2 - 4t + 3 ] > 0 5ivide both sides by 3, t^2 - 4t + 3 > 0 t^2 - 3t - t + 3 > 0 t(t - 3) - 1(t - 3) > 0 (t - 1)(t - 3) > 0 t - 1 > 0 and t - 3 > 0 t > 1 and t > 3 t = { 1 < t < 3: t > 3 } e) Total distance traveled during the first 4s = f(4) f(t) = t^3 - 6t^2 + 9t f(4) = (4)^3 - 6(4)^2 + 9(4) = 64 - 6(16) + 36 = 64 - 96 + 36 = 64 - 60 = 4 ft f(4) = 4ft
The equation of motion of a particle is s=2t^3-7t^2+4t+1 ?
s=2t^3-7t^2+4t+1 a) velocity, v = ds/dt = 6t^2 - 14t + 4 = 2(3t^2 - 7t + 2) m/s acceleration, a = dv/dt = 12t - 14 = 2(6t - 7) m/s^2 b) a(1) = 12(1) - 14 = -2 m /s^2
The equation of motion of a particle is s = t3 − 3t, where s is in meters and t is in seconds. (Assume t ≥ 0?
s(t) = t^3 − 3t where t ≥ 0 a) velocity, v(t) = s'(t) v(t) = 3*t^(3-1) - 3*1t^(1-1) = 3t^2 - 3t^0 = 3t^2 - 3 v(t) = 3[t^2 - 1] [where t ≥ 0] Acceleration, a(t) = v'(t) = s"(t) Where v(t) = 3t^2 - 3 = 3*2t^(2-1) - 3*0t^(0-1) = 6t - 0 = 6t a(t) = 6t [where t ≥ 0] b) accleration after 4s = a(4) a(4) = 6(4) = 24 a(4) = 24m/s^2 c) acceleration when velocity, v(t) = 0 When v(t) = 0 find t, When v(t) = 0, 3[t^2 - 1] = 0 Divide both sides by 3, t^2 - 1 = 0 t^2 = 1 t = ± √[ 1/2 ] t = ± 1/√2 = ±(√2)/2 s Since t ≥ 0, t = +(√2)/2 s t = (√2)/2 s Now, find a(√2/2), a(t) = 6t a((√2)/2) = 6[(√2)/2] = 6/2 [√2] = 3√2 m/s^2
The Position of a particle is given by the Equation s=F(t)= t^3-6t^2=9t?
The Velocity is V=3t^2-12t+9 The question is: How do you draw the diagram in motion? Which equation do I use? How do you find the total distance traveled by the particle during the first 5 seconds
The acceleration of a particle, a m/s^2 moving in a straight line at time t seconds is given by a = t + 1. How do I find the formula for velocity and displacement given that s=0 and v=8 when t=0?
Acceleration is given by:[math]a=\dfrac{dv}{dt}\implies t+1=\dfrac{dv}{dt}\implies[/math] [math]dv= (t+1)dt[/math]Integrating both sides w.r.t. dt,[math]\int (t+1)dt=\int dv[/math][math]\dfrac{t^2}{2}+t+c=v... (1)[/math](c is the constant of integration when an indefinite integral is evaluated )According to the condition,at t=0,v=8.Substituting t=0,v=8 in (1),c=8.So the equation (1) becomes[math]v=\dfrac{t^2}{2}+t+8 ... (2)[/math]To obtain the expression for displacement,write v=ds/dt.Thus (2) can be written as:[math]\dfrac{ds}{dt}=\dfrac{t^2}{2}+t+8\implies ds=(\dfrac{t^2}{2}+t+8)dt[/math]Again Integrating both sides w.r.t dt,[math]\int ds=\int(\dfrac{t^2}{2}+t+8)dt[/math][math]s=\dfrac{t^3}{6}+\dfrac{t^2}{2}+8t+c... (3)[/math](c is the constant of integration )According to the condition at t=0,s=0.Putting t=0,s=0 in eqn (3)c=0.Thus eqn (3) becomes[math]s=\dfrac{t^3}{6}+\dfrac{t^2}{2}+8t[/math]This the equation for displacement.Hope it helps!!