Acceleration/veloctiy/displacement/time question

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In summary, the problem involves two runners, Michael and Robert, in a long distance race. Michael is running at 3.8 m/s and is 75 m behind Robert, who is running at a constant velocity of 4.2 m/s. Given Michael's acceleration of 0.15 m/s2, the question is how long it will take him to catch Robert. To solve this, two equations are needed: t = d / 4.2m/s and d = (3.8m/s)(t) + (1/2)(0.15m/s2)(t2), where d represents the distance traveled by Michael when he catches up to Robert, taking into account Robert's initial lead of
  • #1
phongbui
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Ok i just started gr 11 physics, and I am wondering how would you tackle this problem

In a long distance race, Michael is running at 3.8 m/s and is 75 m behind robert, who is running at a constant velocity of 4.2 m/s. If Michael accelerates at 0.15 m/s2, how long will it take him to catch Robert?

Im assuming in this question you need 2 equations to find the unknown, I don't know how to make an equation with robert being ahead 75m.. :(

Possible formulas that could be used are

V = d/t

d = v1 x t + 1/2(a)t2 <-- (time squared)

a = vf - vi / t <--- final velocity - intial velocity

d = vi + vf /2 (t)

Please keep things simple ^^ as I've just started gr 11 physics :)

Thanks for the help
 
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  • #2
Welcome to PF.
You need two equations to solve this problem.
In time t Robert move a distance x. The in the same time what is the distance moved by Michael when he meets Robert?
 
  • #3
Thanks for the quick reply

hmm... so the first equation would be t = d / 4.2m/s

than the 2nd part... would be d = (3.8m/s)(t) + (1/2)(0.15m/s2)(t2)

is this right so far?
 
  • #4
phongbui said:
Thanks for the quick reply

hmm... so the first equation would be t = d / 4.2m/s

than the 2nd part... would be d = (3.8m/s)(t) + (1/2)(0.15m/s2)(t2)

is this right so far?

The first one is
d = 4.2 m/s*t
And in the second one the displacement is d + 75 m.
 
  • #5
Ohhh... Thanks very much
 

Related to Acceleration/veloctiy/displacement/time question

1. What is the difference between acceleration and velocity?

Acceleration refers to the rate at which an object's velocity changes over time, while velocity is the speed and direction of an object's motion at a particular point in time.

2. How is acceleration calculated?

Acceleration can be calculated by dividing the change in velocity by the change in time. This can also be represented as the slope of a velocity vs. time graph.

3. What is displacement?

Displacement is the distance and direction of an object's change in position from its starting point. It is a vector quantity, meaning it has both magnitude and direction.

4. How is displacement related to velocity and time?

Displacement is directly related to an object's velocity and the time it takes to travel that velocity. The formula for displacement is displacement = velocity x time.

5. Can an object have a constant velocity and non-zero acceleration?

No, an object cannot have a constant velocity and non-zero acceleration. This would violate the definition of acceleration, which requires a change in velocity over time. If an object has a constant velocity, its acceleration is 0.

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