SOLUTION: 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) mete

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.
Found 2 solutions by Shin123, ikleyn:
Answer by Shin123(626)   (Show Source): You can put this solution on YOUR website!
The acceleration function is . By definition, acceleration is the rate of change in velocity, aka the derivative of velocity. So taking the integral gives us the velocity as , for some constant c. We know the initial velocity is 2 m/s, aka v(0)=2. This means that c=2, so . This allows us to answer a), 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 , where d is a constant. Since we're only considering displacement, we can assume the particle starts at position 0, so . We can now answer a), getting meters of displacement.
Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.

        In the post by tutor @Shin, the answer to question (a) is correct,
        while the answer to question (b) is INCORRECT.

The correct answer to question (b) is the displacement of the particle in the fifth second is s(5) - s(4), where s(t) = - + 2t. So, s(4) = - + 2*4 = -18.6667 m (rounded), s(5) = - + 2*5 = -23.3333 m (rounded), and the displacement in the fifth second is s(5) - s(4) = -23.3333 - (-18.6667) = -4.6666 m (rounded). ANSWER to (b) is -4.6667 m (rounded).

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