# 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 the car coasts before it stops. Express your answer to two significant figures and include the appropriate units.

2) Calculate the time it takes to stop. Express your answer to two significant figures and include the appropriate units.

3) Calculate the distance it travels during the first second. Express your answer to two significant figures

4) Calculate the distance it travels during the fifth second.

Express your answer to two significant figures

### 5 Answers

- 2 months agoBest Answer
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

- oubaasLv 72 months ago
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*1-a/2*1^2 = 67/3.6-0.225 = 18 m

d1 = (67/3.6+67/3.6-0.45*1)/2*1 = 18 m

4) Calculate the distance d2 it travels during the fifth second to two significant figures

V4th = (67/3.6-0.45*4) = 16.81 m/sec

V5th = (67/3.6-0.45*5) = 16.36 m/sec

d2 = (16.81+16.36)/2*1 = 17.0 m

- Jim MoorLv 72 months ago
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

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- Anonymous2 months ago
Should have paid attention in class.