Kinematics - A maths
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A particle moving in a straight line passes a fixed point A with a velocity of -7ms-1/ The acceleration (after passing A) is given by a = 8 - 2t for 0<t<T. On reaching its maximum speed at T seconds, the particle begins to decelerate at a constant rate of 1.5ms-2 until it comes to rest at the point B.
Find the total distance traveled by the particle from A to B.
Drew out a lot of the motion graphs but kept getting the wrong answer. I found that the max speed is 9 ms-1 and T = 4. The 2 points where particle makes U-turn are at time 1 and 7 second. Just to ask, at which point does particle come to a REST? Because it seems like at the 7th second, speed is already at 0ms-1 so, how does particle continue its deceleration?
Immediate help needed. Thank you very much!
Forgot to add, answer's 48 1/3.
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Hi,
At the maximum speed, a = 0, so you are right that T = 4.
Next, we need to find v by integration, i.e. v = 8t - t^2 + c.
When t = 0, v = -7, so v = 8t - t^2 - 7.
Sketch a graph of v against t. This is an inverted u-shaped curve.
We only consider the region of the graph between t = 0 and t = 4.
During this period of time, distance travelled
= - integral { 0, 1 } v dt + integral { 1, 4 } v dt --- (1)
Beyond t = 4, we need to use distance travelled = ut + 1/2 at^2, because we are told that the previous acceleration formula applies only for 0 < t < T.
From the maximum velocity of 9 m/s, it will take 6s to drop to 0 m/s at the acceleration of -1.5 m/s^2.
Now, distance travelled = (9)(6) + 1/2 (-1.5)(6^2) = 27 --- (2)
Hence total distance = (1) + (2).
Thanks!
Cheers,
Wen Shih -
That helps! Thanks!
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Hi,
Wah lau, difficult for O-level students leh! Many levels of thinking...
Cheers,
Wen Shih -
just reply lah, this is not a letter
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Hi,
Problem already solved in the 2nd post. I'm just making a comment :)
Cheers,
Wen Shih -
Originally posted by wee_ws:
Hi,
Wah lau, difficult for O-level students leh! Many levels of thinking...
Cheers,
Wen ShihDamn funny
Hi is followed by wah lau
Formal + informal