Practice Physics Question

[MissionStanford]

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A car accelerates steadily over 10 seconds from a stationary position to 25 m/s. The car maintains its velocity for 20 seconds, then decelerates steadily back to a stationary position in 5 seconds. What is its total displacement from start to finish?

Here's what I did:
d = (1/2)at^2 = (1/2)(25/10)(10)^2 = 125
d = vt = (25)(20) = 500
d = (1/2)(-25/5)(5)^2 = -62.5

That sums up to 562.5, but I quickly realized that even while decelerating the car moves in the same direction, so it should actually be positive 62.5 at the end, which leads to a total of 687.5, the right answer. Why am I getting a negative value for displacement on the third part though, if that's not right?

 

[MissionStanford]

Actually, I think I figured it out. At the end it should be the full equation:

d = v0t + (1/2)at^2
d = (25)(5) + (1/2)(-5)(25)
d = 125 - 62.5
d = 62.5m

 

[gettheleadout]

Yes, the displacement equation also contains v_0 * t.

 

[Temperature101]

A car accelerates steadily over 10 seconds from a stationary position to 25 m/s. The car maintains its velocity for 20 seconds, then decelerates steadily back to a stationary position in 5 seconds. What is its total displacement from start to finish?

Here's what I did:
d = (1/2)at^2 = (1/2)(25/10)(10)^2 = 125
d = vt = (25)(20) = 500
d = (1/2)(-25/5)(5)^2 = -62.5

That sums up to 562.5, but I quickly realized that even while decelerating the car moves in the same direction, so it should actually be positive 62.5 at the end, which leads to a total of 687.5, the right answer. Why am I getting a negative value for displacement on the third part though, if that's not right?

That problem almost tripped me up yesterday on mcatquestion.com...It took me ten minutes to realize that the equation should be d = VoT + 1/2 aT^2 and that Vo is 25 m/s. I was using just d = 1/2at^2 like you...In the end I figured it out. But I cant afford to spend all that time on a simple question like that in the mcat.