Question

Displacement & Constant Acceleration

A driver in a car traveling at a speed of 21.8m/s sees a cat 101 m away on the road. How long will it take for the car to accelerate uniformly to a stop in exactly 99 m.

Answer 1

To solve this question you need use the equations of motion. In particular, the equation: \[s=\left( \frac{ v + u }{ 2 } \right)t\] is what we will use. Where s is the displacement (ie. 99m), v is the final velocity, u is the initial velocity, and t is time. Therefore, \[99 = \left( \frac{ 0 + 21.8 }{ 2 } \right)t\]. Solving for this: \[t = \frac{ 99 }{ 10.9 } = 9.08\approx 9.1\], where t is in seconds. Also, I know this is just knit picking but the car is not accelerating to a stop, rather it is decelerating (negative acceleration).

Answer 2

@carson889 can I show you what I did?

Answer 3

Sure, go at it.

Answer 4

a = ?
Vi = 21.8
Vf = 99
t = ?
x = 101

I was trying to use the formula that would give me a or t.
Looks like either...

x=Vi*t+1/2a(t)^2

or

Vf^2=Vi^2+2a*x


???
Then I got stuck.

Answer 5

@carson889

Answer 6

my book says the answer is -7.5m/s; 19m

Answer 7

If the question is asking for the time it takes to stop in 99 meters, then the answers in the book don't quite make sense, as they are measures of velocity and displacement. Ensure you are looking at the right answer or question.
Next, the equations you gave were mostly right.
a = ? We don't know the value of acceleration yet so a = ?
Vi = 21.8 , correct
Vf = 99 , this is the final velocity. Assuming the car is coming to a stop our final velocity will be zero. Thus Vf should equal 0.
t = ?
x = 101, x is the displacement. We know the cat is 101m ahead, however the question asks us to find how long it takes to stop in 99m. Thus x should equal 99m.

Answer 8

ok

Answer 9

So I can use the second equation

Vf^2=Vi^2+2a*x

Answer 10

0=21.8^2+2a*99

Answer 11

The two formulas you posted:

x=Vi*t+1/2a(t)^2
or
Vf^2=Vi^2+2a*x

don't exactly work in this case. For the first one, we would need to already know either the time or acceleration to complete the equation. We can use the second one to find the deceleration (negative acceleration) and then plug that into the first equation to find time. However, use the equation I first mentioned will be a much more simple solution. Just for kicks though:

\[V_{f}^{2} = V_{i}^{2} + 2ax\]
0 = 21.8^2 + 2a*(99)
\[\frac{ 21.8^{2} }{ 2(99) }=a\]
a = 2.4 m/s^2

First equation:
x = Vi*t + 1/2(a*(t^2))
Since Vi is zero the first part doesn't matter, thus.
\[\frac{ 2x }{ a} = t ^{2}\]
\[\frac{ 99*2 }{ 2.4 } = t ^{2}=82.5,t=\sqrt{82.5}=9.08\approx9.1\]

Answer 12

Technically a should equal -2.4 m/s^2. However, when you plug it into the first equation it is squared, and as we know a negative multiplied by a negative is positive.

Answer 13

The equation you used is not on my list.

Answer 14

I only have 4 to choose from.

Answer 15

Then it looks as if they want you to do it the longer way by using the two equation you provided. You first find the acceleration and then plug that into the other equation to find the time. The first equation I used is still valid as it is just one of the equations of motion, it will give you the same time as using the two equations.