a particle moves in a straight line so that , $t$ s after passing through a fixed point $O$ , its velocity $v$ in $\frac{m}{s}$, is given by $$v = 2t -11-\left(\frac{6}{(t+1)}\right)$$ Find the acceleration of the particle when it is at instantaneous rest.
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1Are you familiar with the interpretation of acceleration as the derivative of velocity with respect to time? – hardmath Oct 30 '18 at 14:57
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yes...@hardmath please guide me further. thank you – Dexter Khoo Jun Ming Oct 30 '18 at 14:58
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2"Shouldn't acceleration be $0$ when velocity is $0$?" No, not necessarily. Throw a ball straight up into the air. Ignoring the moment it leaves your hand from that point on the acceleration will be only due to gravity (and wind resistance) and will be (practically) constant throughout it's motion. At the peak of the arc the ball will have $0$ velocity as it changes from going up to going down but the acceleration is still the constant $-9.81$ meters per second squared. – JMoravitz Oct 30 '18 at 14:59
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thanks @JMoravitz but when it is at its peak it wouldn't have any direction to move , so it would have no acceleration? – Dexter Khoo Jun Ming Oct 30 '18 at 15:01
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As for your specific question., you write $v=2t-11=(6/(t+1))$. This makes no sense. You have two equals signs and what you have equated do not equal eachother except perhaps at two values of $t$. You must have some sort of typo here. – JMoravitz Oct 30 '18 at 15:01
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@JMoravitz there is an error wait please – Dexter Khoo Jun Ming Oct 30 '18 at 15:03
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3"When it is at its peak it wouldn't have any direction to move, so it would have no acceleration?" Between this comment and your original question you seem to be holding on to a very naive understanding of what acceleration means and do not know the proper definition. I recommend that you forget everything you thought you knew about speed, velocity, acceleration, etc... and instead approach it fresh from the definitions. Velocity is the change in position over time. Acceleration is the change in velocity over time. – JMoravitz Oct 30 '18 at 15:05
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understood @JMoravitz thanks for your guidance – Dexter Khoo Jun Ming Oct 30 '18 at 15:08
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so how do I find the acceleration after I differentiated the equation? – Dexter Khoo Jun Ming Oct 30 '18 at 15:16
1 Answers
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In the following picture your problem is solved within Mathematica.The most important feature is the plot. It shows that for small positive $t$ the velocity is negative, hence the moving point moves downwards (say). But the velocity (a signed quantity!) is steadily rising, and after about $6$ seconds it is exactly zero, then becomes positive, which means that from then on the moving point begins moving upwards. Now this steadily rising of $v(t)$ is probably caused by some extraneous force. In any case the intensity of this rising of $v(t)$ is called the acceleration of the moving point, and is measured by $a(t):=v'(t)$.
Christian Blatter
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