My attempt seems right? (Kinematics)

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A racing car accelerates from 10 m/s to 50 m/s over a distance of 60 m, leading to a discussion about the time taken for this change. The initial calculations suggest a time of 2 seconds, but the textbook indicates the answer is 4 seconds. The confusion arises from the application of kinematic equations, with participants confirming that constant acceleration leads to a linear velocity slope. The correct approach involves using the kinematic equation to derive the acceleration and subsequently the time, confirming that 2 seconds is indeed the calculated result based on the provided parameters. Ultimately, the consensus is that the calculations support a time of 2 seconds, despite the textbook's claim.
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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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