document.write( "Question 1196122: A particle moving in a straight line has initial velocity of 2 m/s at a point O on the line. The particle moves so that its acceleration is t seconds later is given by (2t-6) meter per second squared. Find the (a) Velocity when t= 5 seconds (b) Displacement of the particle in the fifth seconds. \n" ); document.write( "

Algebra.Com's Answer #844768 by Shin123(626) $\"\"$ $\"About$

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The acceleration function is $\"a%28t%29=2t-6\"$ . By definition, acceleration is the rate of change in velocity, aka the derivative of velocity. So taking the integral gives us the velocity as $\"v%28t%29=t%5E2-6t%2Bc\"$ , for some constant c. We know the initial velocity is 2 m/s, aka v(0)=2. This means that c=2, so $\"v%28t%29=t%5E2-6t%2B2\"$ . This allows us to answer a), $\"v%285%29=5%5E2-6%2A5%2B2=25-30%2B2=-3\"$ meters per second. The velocity is the rate of change in position, aka the velocity is the derivative of the position. Taking the integral again gives us $\"s%28t%29=t%5E3%2F3-3t%5E2%2B2t%2Bd\"$ , where d is a constant. Since we're only considering displacement, we can assume the particle starts at position 0, so $\"s%28t%29=t%5E3%2F3-3t%5E2%2B2t\"$ . We can now answer a), getting $\"s%285%29=5%5E3%2F3-3%2A5%5E2%2B2%2A5=-70%2F3\"$ meters of displacement. \n" ); document.write( "

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