My attempt seems right? (Kinematics)

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Homework Help Overview

The problem involves a racing car accelerating from 10 m/s to 50 m/s over a distance of 60 m, with participants exploring how to calculate the time taken for this acceleration. The subject area is kinematics, specifically focusing on equations of motion under constant acceleration.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants attempt to apply kinematic equations to find the time taken, with some expressing uncertainty about their calculations. Questions arise regarding the validity of the provided answers and the assumptions made about constant acceleration.

Discussion Status

The discussion is ongoing, with various interpretations of the problem being explored. Some participants suggest that the original poster's calculations may be incorrect, while others question the reliability of the textbook answers. There is no explicit consensus on the correct approach or solution yet.

Contextual Notes

Participants note potential discrepancies in the textbook answers and discuss the implications of constant acceleration on the calculations. There is also mention of physical constraints regarding the time variable in the context of the problem.

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A racing car traveling with constant acceleration increases its speed from 10m/s to 50m/s over
a distance of 60 m. How long does this take?

A. 2.0 s
B. 4.0 s
C. 5.0 s
D. 8.0 s
E. The time cannot be calculated since the speed is not constant

I did:

x - xo = 0.5 (Vo + V)(t)

Which is:

60 = 0.5 (60) (t)

Which is:

60 = 30t
t = 2

But, the answer is 4.

What am I doing wrong?
 
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x=v0t+1/2*at^2

a= (v-v0)/t

combine and simplify
x=1/2(v+v0)t

t=(2x)/(v+v0)=2

bad answer I guess.
 
Are you sure? The answers came from my college textbook, but I suppose they could be wrong?
 
Well, a constant acceleration makes for a velocity with a linear slope, and the slope of that velocity is by definition the acceleration (it's not even the mean acceleration as long as acceleration is constant). The slope of the velocity is ∆v/∆t=40/t m/s^2 (since t starts at 0).

Plug this into the kinematic equation, and, of course, you get the same thing as the algebra.
60=10*t+20*t

you can also get a directly
(v^2-v0^2)/(2*x)=a

a=20m/s^2

if you plug this into the quadratic kinematic equation

0=-60+10t+10t^2

t=-3,2 (-3 is non-physical)

Either way, as you've posted the problem, the answer is most definitely 2s.
 

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