Questions: Starting from rest, a particle moving in a straight line has an acceleration of a=(2 t-6) m / s^2, where t is in seconds. What is the particle's velocity when t=6 s, and what is its position when t=11 s?

Solution Steps

• The acceleration \( a(t) = 2t - 6 \) is given.
• Velocity \( v(t) \) is the integral of acceleration with respect to time.
• Integrate \( a(t) \) to find \( v(t) \): \[ v(t) = \int (2t - 6) \, dt = t^2 - 6t + C \]
• Since the particle starts from rest, \( v(0) = 0 \).
• Substitute \( t = 0 \) into the velocity equation to find \( C \): \[ 0 = 0^2 - 6 \times 0 + C \implies C = 0 \]
• Therefore, the velocity function is: \[ v(t) = t^2 - 6t \]

• Substitute \( t = 6 \) into the velocity function: \[ v(6) = 6^2 - 6 \times 6 = 36 - 36 = 0 \, \text{m/s} \]

• Position \( s(t) \) is the integral of velocity with respect to time.
• Integrate \( v(t) = t^2 - 6t \) to find \( s(t) \): \[ s(t) = \int (t^2 - 6t) \, dt = \frac{t^3}{3} - 3t^2 + D \]
• Since the particle starts from rest at the origin, \( s(0) = 0 \).
• Substitute \( t = 0 \) into the position equation to find \( D \): \[ 0 = \frac{0^3}{3} - 3 \times 0^2 + D \implies D = 0 \]
• Therefore, the position function is: \[ s(t) = \frac{t^3}{3} - 3t^2 \]

• Substitute \( t = 11 \) into the position function: \[ s(11) = \frac{11^3}{3} - 3 \times 11^2 \]
• Calculate each term: \[ \frac{11^3}{3} = \frac{1331}{3} \approx 443.67 \] \[ 3 \times 11^2 = 3 \times 121 = 363 \]
• Therefore, the position at \( t = 11 \, \text{s} \) is: \[ s(11) \approx 443.67 - 363 = 80.67 \, \text{m} \]

Final Answer