Solving Uniform Acceleration: Calculating Velocity & Distance

In summary, the conversation is about a person seeking help in determining the value of uniform acceleration, velocities, and total distance traveled for a car that is observed to cover distances of 6m and 7.5m in two consecutive intervals of 1 second each. After some discussion and calculations, it is determined that the acceleration is not 1.5m/s^2 as previously thought, but rather a different kinematics relationship must be used.
  • #1
Bmrboi
4
0
Hi guys,

New here, i have a question which has been doing my head in for days now. Hope someone might know how to answer it.

The question is: A car travels with uniform acceleration from rest and is observed to cover distances of 6m and 7.5m respectively in two consecutive intervals of 1 second each.
Determine;

i) the value of the uniform acceleration,

ii) the velocities at the beginning and end of each of these two seconds,

iii) the total distance traveled from rest till end of the 7.5m distance period.


I can work out part ii) and iii) if i had acceleration, but i cannot seem to work it out.
I have drawn a velocity/time graph but it just confuses me even more.

From my calculations i work a= 1.5m/s2

Any help would be great!

Thanks Bmrboi
 
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  • #2
The acceleration is [itex]\dfrac{dv}{dt}=\dfrac{\Delta v}{\Delta t}[/itex]
 
  • #3
luitzen said:
The acceleration is [itex]\dfrac{dv}{dt}=\dfrac{\Delta v}{\Delta t}[/itex]

Thanks for the reply, are you meaning a= (7.5m-6m)/ (2s-1s)? Which Equals 1.5m/s2?
 
  • #4
Yes. The acceleration is the slope of the velocity/time graph.

To be correct, the speed is not in m, but in m/s.

The first time interval is 6m in 1s or 6m/1s=6m/s. This will get more important when problems become less obvious and you can recognize mistakes by checking for wrong dimensions.
 
  • #5
Yes you are right. I have answers now! thanks!
 
  • #6
Bmrboi said:
Thanks for the reply, are you meaning a= (7.5m-6m)/ (2s-1s)? Which Equals 1.5m/s2?

No, this is not quite right. If your acceleration is constant then you can indeed use this formula, but how have you figured out that the speed at the end of the first second is 6 m/s? If you think for a second you'll realize that you have no basis for thinking that. It would only be true if it uniformly had a speed of 6 m/s for the first second, in which case it would have zero acceleration (which isn't true). You can also prove that 1.5 m/s[itex]^2[/itex] is wrong: that means that every second, the speed grows by 1.5 m/s, by definition. But your claim is that the speed is 6m/s after the end of one second. This is a contradiction.

So you'll have to try again, with a different kinematics relationship.
 

FAQ: Solving Uniform Acceleration: Calculating Velocity & Distance

What is uniform acceleration?

Uniform acceleration refers to a situation in which an object's velocity changes at a constant rate. This means that the object's acceleration remains constant over time, resulting in a linear change in velocity.

How do you calculate velocity in uniform acceleration?

To calculate velocity in uniform acceleration, you can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time interval. Alternatively, you can use the formula v = u + 2as, where s is the displacement of the object.

What is the equation for distance in uniform acceleration?

The equation for distance in uniform acceleration is s = ut + 1/2at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time interval.

How do you solve for acceleration in uniform acceleration?

To solve for acceleration in uniform acceleration, you can rearrange the formula a = (v - u)/t, where v is the final velocity, u is the initial velocity, and t is the time interval. Alternatively, you can use the formula a = (2s - 2ut)/(t^2), where s is the displacement and u is the initial velocity.

Can uniform acceleration be negative?

Yes, uniform acceleration can be negative. This means that the object's velocity is decreasing at a constant rate over time. In this case, the formula for velocity would be v = u - at, and the formula for distance would be s = ut - 1/2at^2.

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