A particle P of mass 1 kg is subjected to a force F = [(3t2 - 4t)i + (6t - 5)j] N. If its velocity at t=3s is (11i + 10j) m/s, find its acceleration and velocity at time t.

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Asked on April 11, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-04-11T11:24:07.398Z

You're on a roll — Given: Mass of particle P, m = 1 kg. Force F = [(3t^2 - 4t)i + (6t - 5)j] N. At t=3 s, velocity v(3) = (11i + 10j) m/s. a) The acceleration of P at time t seconds. Step 1: Use Newton's second law, F = ma. Since the mass m = 1 kg, the acceleration a is equal to the force F. a(t) = F(t) a(t) = (3t^2 - 4t)i + (6t - 5)j m/s^2 The acceleration of P at time t is (3t^2 - 4t)i + (6t - 5)j m/s^2. b) The velocity of P at time t seconds. Step 1: Integrate the acceleration vector to find the velocity vector. v(t) = a(t)\,dt v(t) = [(3t^2 - 4t)i + (6t - 5)j]\,dt v(t) = ( (3t^2 - 4t)\,dt)i + ( (6t - 5)\,dt)j v(t) = (t^3 - 2t^2 + C_1)i + (3t^2 - 5t + C_2)j Let C = C_1 i + C_2 j be the constant of integration. v(t) = (t^3 - 2t^2)i + (3t^2 - 5t)j + C Step 2: Use the given initial condition to find the constant of integration C. We are given that v(3) = (11i + 10j) m/s. Substitute t=3 into the velocity equation: v(3) = (3^3 - 2(3)^2)i + (3(3)^2 - 5(3))j + C v(3) = (27 - 18)i + (27 - 15)j + C v(3) = 9i + 12j + C Equate this to the given velocity at t=3: 11i + 10j = 9i + 12j + C C = (11i + 10j) - (9i + 12j) C = (11 - 9)i + (10 - 12)j C = 2i - 2j Step 3: Substitute C back into the velocity equation. v(t) = (t^3 - 2t^2)i + (3t^2 - 5t)j + (2i - 2j) v(t) = (t^3 - 2t^2 + 2)i + (3t^2 - 5t - 2)j m/s The velocity of P at time t is (t^3 - 2t^2 + 2)i + (3t^2 - 5t - 2)j m/s. c) The kinetic energy of P when it is moving parallel to the vector i. Step 1: Determine the time t when the particle is moving parallel to the vector i. Moving parallel to the vector i means the j-component of the velocity is zero. From part (b), the j-component of v(t) is (3t^2 - 5t - 2). Set this to zero: 3t^2 - 5t - 2 = 0 Solve the quadratic equation for t: t = -(-5) ± sqrt((-5)^2 - 4(3)(-2))2(3) t = 5 ± sqrt(25 + 24)6 t = 5 ± sqrt(49)6 t = (5 ± 7)/(6) Two possible values for t: t_1 = (5 + 7)/(6) = (12)/(6) = 2 s t_2 = (5 - 7)/(6) = (-2)/(6) = -(1)/(3) s Since time must be positive, we use t=2 s. Step 2: Calculate the velocity of P at t=2 s. Substitute t=2 into the velocity equation from part (b): v(2) = ((2)^3 - 2(2)^2 + 2)i + (3(2)^2 - 5(2) - 2)j v(2) = (8 - 8 + 2)i + (12 - 10 - 2)j v(2) = 2i + 0j v(2) = 2i m/s The magnitude of the velocity is |v(2)| = sqrt(2^2 + 0^2) = sqrt(4) = 2 m/s. Step 3: Calculate the kinetic energy. The kinetic energy KE is given by KE = (1)/(2)mv^2. Given m=1 kg and v=2 m/s. KE = (1)/(2)(1 kg)(2 m/s)^2 KE = (1)/(2)(1)(4) KE = 2 J The kinetic energy of P when it is moving parallel to the vector i is 2 J. d) Determine the velocity of P relative to Q at time t=2. Step 1: Identify the velocity of particle Q. Particle Q moves with velocity v_Q = (3i + 4j) m/s. Step 2: Identify the velocity of particle P at t=2 s. From part (c), we found v_P(2) = 2i m/s. Step 3: Calculate the velocity of P relative to Q. The velocity of P relative to Q is v_P/Q = v_P - v_Q. v_P/Q = (2i) - (3i + 4j) v_P/Q = (2 - 3)i + (0 - 4)j v_P/Q = -1i - 4j m/s The velocity of P relative to Q at time t=2 is -i - 4j m/s. What's next?

Accepted Answer · 2026-04-11T11:24:07.398Z

You're on a roll — Given: Mass of particle P, m = 1 kg. Force F = [(3t^2 - 4t)i + (6t - 5)j] N. At t=3 s, velocity v(3) = (11i + 10j) m/s. a) The acceleration of P at time t seconds. Step 1: Use Newton's second law, F = ma. Since the mass m = 1 kg, the acceleration a is equal to the force F. a(t) = F(t) a(t) = (3t^2 - 4t)i + (6t - 5)j m/s^2 The acceleration of P at time t is (3t^2 - 4t)i + (6t - 5)j m/s^2. b) The velocity of P at time t seconds. Step 1: Integrate the acceleration vector to find the velocity vector. v(t) = a(t)\,dt v(t) = [(3t^2 - 4t)i + (6t - 5)j]\,dt v(t) = ( (3t^2 - 4t)\,dt)i + ( (6t - 5)\,dt)j v(t) = (t^3 - 2t^2 + C_1)i + (3t^2 - 5t + C_2)j Let C = C_1 i + C_2 j be the constant of integration. v(t) = (t^3 - 2t^2)i + (3t^2 - 5t)j + C Step 2: Use the given initial condition to find the constant of integration C. We are given that v(3) = (11i + 10j) m/s. Substitute t=3 into the velocity equation: v(3) = (3^3 - 2(3)^2)i + (3(3)^2 - 5(3))j + C v(3) = (27 - 18)i + (27 - 15)j + C v(3) = 9i + 12j + C Equate this to the given velocity at t=3: 11i + 10j = 9i + 12j + C C = (11i + 10j) - (9i + 12j) C = (11 - 9)i + (10 - 12)j C = 2i - 2j Step 3: Substitute C back into the velocity equation. v(t) = (t^3 - 2t^2)i + (3t^2 - 5t)j + (2i - 2j) v(t) = (t^3 - 2t^2 + 2)i + (3t^2 - 5t - 2)j m/s The velocity of P at time t is (t^3 - 2t^2 + 2)i + (3t^2 - 5t - 2)j m/s. c) The kinetic energy of P when it is moving parallel to the vector i. Step 1: Determine the time t when the particle is moving parallel to the vector i. Moving parallel to the vector i means the j-component of the velocity is zero. From part (b), the j-component of v(t) is (3t^2 - 5t - 2). Set this to zero: 3t^2 - 5t - 2 = 0 Solve the quadratic equation for t: t = -(-5) ± sqrt((-5)^2 - 4(3)(-2))2(3) t = 5 ± sqrt(25 + 24)6 t = 5 ± sqrt(49)6 t = (5 ± 7)/(6) Two possible values for t: t_1 = (5 + 7)/(6) = (12)/(6) = 2 s t_2 = (5 - 7)/(6) = (-2)/(6) = -(1)/(3) s Since time must be positive, we use t=2 s. Step 2: Calculate the velocity of P at t=2 s. Substitute t=2 into the velocity equation from part (b): v(2) = ((2)^3 - 2(2)^2 + 2)i + (3(2)^2 - 5(2) - 2)j v(2) = (8 - 8 + 2)i + (12 - 10 - 2)j v(2) = 2i + 0j v(2) = 2i m/s The magnitude of the velocity is |v(2)| = sqrt(2^2 + 0^2) = sqrt(4) = 2 m/s. Step 3: Calculate the kinetic energy. The kinetic energy KE is given by KE = (1)/(2)mv^2. Given m=1 kg and v=2 m/s. KE = (1)/(2)(1 kg)(2 m/s)^2 KE = (1)/(2)(1)(4) KE = 2 J The kinetic energy of P when it is moving parallel to the vector i is 2 J. d) Determine the velocity of P relative to Q at time t=2. Step 1: Identify the velocity of particle Q. Particle Q moves with velocity v_Q = (3i + 4j) m/s. Step 2: Identify the velocity of particle P at t=2 s. From part (c), we found v_P(2) = 2i m/s. Step 3: Calculate the velocity of P relative to Q. The velocity of P relative to Q is v_P/Q = v_P - v_Q. v_P/Q = (2i) - (3i + 4j) v_P/Q = (2 - 3)i + (0 - 4)j v_P/Q = -1i - 4j m/s The velocity of P relative to Q at time t=2 is -i - 4j m/s. What's next?

A particle P of mass 1 kg is subjected to a force F = [(3t2 - 4t)i + (6t - 5)j] N. If its velocity at t=3s is (11i + 10j) m/s, find its acceleration and velocity at time t.

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A particle P of mass 1 kg is subjected to a force F = [(3t2 - 4t)i + (6t - 5)j] N. If its velocity at t=3s is (11i + 10j) m/s, find its acceleration and velocity at time t.

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