How Do You Sketch a Velocity-Time Graph for Constant Velocity and Acceleration?

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To sketch a velocity-time graph for a particle moving with a constant velocity of 5 m/s for 2 seconds followed by a constant acceleration of -2 m/s² for 8 seconds, the initial segment will be a horizontal line at 5 m/s, and the second segment will slope downwards due to the negative acceleration. The final velocity after 10 seconds is calculated to be -10 m/s, but this is incorrect; the correct final velocity should be determined using the acceleration formula. The total distance covered is reported as 44.5 m, which needs verification against the correct calculations. Both velocity and acceleration are vectors, and understanding their directional properties is crucial for accurate graphing and calculations.
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Hi guys - i need some help with a really simple problem. I've just started doing physics in college, and I've had a homework question that just doesn't work out for me (I also have the answers).

A particle moves in a straight line with a constant velocity of 5m/s for 2 seconds. It then moves with a constant acceleration of -2m/s^2 for 8 seconds. Sketch a velocity-time graph for the interval of 10 seconds and find:

a) the final velocity
b)the total distance covered by the particle
c)the increase in displacement of the particle

i'm getting -10m/s for a - then 44.5m for b.

Could someone tell me if I'm atleast heading in the right direction? (no pun intended)
 
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iamBevan said:
Hi guys - i need some help with a really simple problem. I've just started doing physics in college, and I've had a homework question that just doesn't work out for me (I also have the answers).

A particle moves in a straight line with a constant velocity of 5m/s for 2 seconds. It then moves with a constant acceleration of -2m/s^2 for 8 seconds. Sketch a velocity-time graph for the interval of 10 seconds and find:

a) the final velocity
b)the total distance covered by the particle
c)the increase in displacement of the particle

i'm getting -10m/s for a - then 44.5m for b.

Could someone tell me if I'm at least heading in the right direction? (no pun intended)

If the initial velocity is 5 [given] and the final velocity is -10 [your calculation] then the change is -15.

AN acceleration of -2 for 8 seconds does not give -15 .
 
If I know my vector is -2m/s^2 then - how will I work out the final velocity of that vector from that?
 
iamBevan said:
If I know my vector is -2m/s^2 then - how will I work out the final velocity of that vector from that?

Why did you say your vector was -2 m/s2?

Surely you meant your acceleration was -2 m/s2.
 
Isn't the vector describing an acceleration?
 
iamBevan said:
Isn't the vector describing an acceleration?

Both velocity and acceleration are vectors
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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