Aleksander Krol
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A particle P is moving along the x-axis with constant deceleration 2.5 m s − 2. At time t = 0 s, P passes through the origin
with velocity 20 m s − 1 in the direction of x increasing. At time t = 12 s, P is at the point A. Find
a the distance OA,
b the total distance P travels in 12 s.
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Last edited by Aleksander Krol; 1 month ago
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Aleksander Krol
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(Original post by Aleksander Krol)
A particle P is moving along the x-axis with constant deceleration 2.5 m s − 2. At time t = 0 s, P passes through the origin
with velocity 20 m s − 1 in the direction of x increasing. At time t = 12 s, P is at the point A. Find
a the distance OA,
b the total distance P travels in 12 s.
Name:  Screenshot_20210725-085515_Drive.jpg
Views: 18
Size:  67.6 KB
how do they know particle P travels a bit far from A and then returns to A?
i mean the question doesn't mentions anything like that.
Last edited by Aleksander Krol; 1 month ago
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mqb2766
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What do you get when you sketch the velocity - time and displacement - time graphs for the particle?
It should be reasonably clear from the info in the question, but if you need to do a sketch based on the suvat equations then just do it.
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Anonbro1
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(Original post by Aleksander Krol)
how do they know particle P travels a bit far from A and then returns to A?
i mean the question doesn't mentions anything like that.
The speed is initially 20m/s and it is reducing at the rate of 2.5 m/s^2. In 8 seconds it will stop. Then the acceleration is in the opposite direction.
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Aleksander Krol
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(Original post by Anonbro1)
The speed is initially 20m/s and it is reducing at the rate of 2.5 m/s^2. In 8 seconds it will stop. Then the acceleration is in the opposite direction.
oh okay i do understand it now. so the particle's v=0 ms^-1 in 8 secs and the question states at 12s the particle is at A so it must have returned to A after 8 secs that's how it's at point A after 12 secs. thanks!!
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