Yahoo Answers: Answers and Comments for A car traveling 67 km/h slows down at a constant 0.45 m/s2 just by "letting up on the gas."? [Physics]
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From Anonymous
enUS
Sun, 22 Sep 2019 04:53:06 +0000
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Yahoo Answers: Answers and Comments for A car traveling 67 km/h slows down at a constant 0.45 m/s2 just by "letting up on the gas."? [Physics]
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From oubaas: A car traveling 67 km/h slows down at a consta...
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Sun, 22 Sep 2019 06:54:56 +0000
A car traveling 67 km/h slows down at a constant 0.45 m/s2 just by "letting up on the gas."
1) Calculate the distance d the car coasts before it stops to two significant figures and include the appropriate units.
d =V^2/2a = 67^2/(3.6^2*0,9) = 380 m
2) Calculate the time t it takes to stop to two significant figures and include the appropriate units.
t = V/a = 67/(3.6*0.45) = 41 sec
3) Calculate the distance d1 it travels during the first second to two significant figures
d1 = V*1a/2*1^2 = 67/3.60.225 = 18 m
d1 = (67/3.6+67/3.60.45*1)/2*1 = 18 m
4) Calculate the distance d2 it travels during the fifth second to two significant figures
V4th = (67/3.60.45*4) = 16.81 m/sec
V5th = (67/3.60.45*5) = 16.36 m/sec
d2 = (16.81+16.36)/2*1 = 17.0 m

From lunchtime_browser: This is just using standard kinematics. I use...
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Sun, 22 Sep 2019 06:16:44 +0000
This is just using standard kinematics. I use the suvat symbols. s = displacement; u = initial velocity; v = final velocity; a = acceleration; t = time. You might use different symbols, {such as vi instead of u} but you should recognise the form of the equations.
The first issue here is that we need to be consistent with units by converting 67 km/h into m/s.
There are 3600 seconds in an hour, so
67 km/h = (67000 / 3600) = 18.61 m/s
(1) What do we know here?
s = ? {That's what we want to know}
u = 18.61 m/s
v = 0 m/s {It's coasted to a stop}
a = 0.45 m/s² {It's slowing down}
t = {Don't know}
Equation choice:
v² = u² + 2as
Substituting known values
0 = 18.61² + 2 * 0.45 * s
0.9s = 346.373
s = 384.86 m
stopping distance = 380 m {2sig figs}
(2) Time to stop we can use another of the standard equations:
v = u + at
with our values:
0 = 18.61 + (0.45)t
0.45t = 18.61
t = 41.35 s
Stopping time = 41 seconds {2sig figs}
(3) Let's draw up our list of knowns again.
s = ? {That's what we want to know}
u = 18.61 m/s
v = {Don't know}
a = 0.45 m/s² {It's slowing down}
t = 1 second
Another of our standard equations...
s = ut + ½at²
values:
s = 18.61 * 1 + ½ *  0.45 * 1²
s = 18.61  0.225 = 18.385
Distanced travelled in first second = 18 m {2sig figs}
(4)
(i) Distance travelled in first five seconds.
s = ut + ½at²
values:
s = 18.61 * 5 + ½ *  0.45 * 5²
s = 93.05  5.625 = 87.425
(ii) Distance travelled in first FOUR seconds.
s = ut + ½at²
values:
s = 18.61 * 4 + ½ *  0.45 * 4²
s = 77.44  3.6 = 70.84 m
So distance travelled during fifth second
= distance travelled in first five seconds  distance travelled in first four seconds
= 87.425  70.84
= 16.585 m
= 17 m to 2 sig figs

From Jim Moor: y(t) = ½at² +v₀t + h₀. <<<< memori...
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https://answers.yahoo.com/question/index?qid=20190922045306AATHM0Y
Sun, 22 Sep 2019 05:19:32 +0000
y(t) = ½at² +v₀t + h₀. <<<< memorize this. it's used extensively!!!
y is the distance
a is the acceleration
t is time and must agree with the time units of the other factors
v₀ is the initial velocity, 'from rest' means v₀ = 0
h₀ is initial height or position

From SumDude: Not my vehicles. They roll, and roll, and roll.
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Sun, 22 Sep 2019 05:59:31 +0000
Not my vehicles. They roll, and roll, and roll.

From Anonymous: Should have paid attention in class.
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Sun, 22 Sep 2019 05:02:26 +0000
Should have paid attention in class.