Issues finding time in acceleration problem

[bracing]

Homework Statement

You are driving to the grocery store at 16 m/s. You are 130 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration.

x1 = 0 m
v1 = 16 m/s
t1 = 0 s
x2 = ?
v2 = 16 m/s
t2 = 0.50s
x3 = 130 m
v3 = 0 m/s
t3 = ?

Homework Equations

(a) How far are you from the intersection when you begin to apply the brakes?
(b) What acceleration will bring you to rest right at the intersection?
(c) How long does it take you to stop?

The Attempt at a Solution

for (a)
(x2-x1) = ((v1+v2)/2)*(t2-t1)
(x2-0m)=((16m/s+16m/s)/2)*(0.50s-0s)
x2=(32m/s/2)*0.50s
x2=16m/s*0.50s
x2 = 8m

for (c)
(x3-x2) = ((v2+v3)/2)*delta t
122m=(16m/s/2)*delta t
16s=delta t

for (b)
a=delta v/delta t
a=-16m/s/16s
a=-1 m/s^2

I know (a) is correct, where I am having issues is computing the final time so I can then obtain the acceleration.

Thanks,

Bracing

 

[collinsmark]

Hello Bracing,

Welcome to Physics Forums!

Homework Equations

(a) How far are you from the intersection when you begin to apply the brakes?
(b) What acceleration will bring you to rest right at the intersection?
(c) How long does it take you to stop?

These are really part of the problem statement, not the relevant equations. But now I'm just being nitpicky...

The Attempt at a Solution

for (a)
(x2-x1) = ((v1+v2)/2)*(t2-t1)
(x2-0m)=((16m/s+16m/s)/2)*(0.50s-0s)
x2=(32m/s/2)*0.50s
x2=16m/s*0.50s
x2 = 8m

You have calculated that the car travels 8 m in 0.5 seconds, which is fine. But the problem statement is asking you how far away the car is from the intersection. (But I see from below that you have already figured out it is 122 m. If you wanted to get that value from your equation directly, you could do it by putting in the 130 m into your (x2 -x1) term somewhere.)

for (c)
(x3-x2) = ((v2+v3)/2)*delta t
122m=(16m/s/2)*delta t
16s=delta t

Be careful of your math. 122/8 isn't exactly 16.

for (b)
a=delta v/delta t
a=-16m/s/16s
a=-1 m/s^2

Your approach is sound. But you'll need to correct that previous mistake first (the time isn't exactly 16 seconds)

 

[bracing]

16s is rounded to significant figures as my instructor instructed the class on how he wanted it. I personally think I'm losing a bit by doing it but he continually stresses that we do.

I am dumb-founded over this as every example given in class had a stated acceleration, then my homework questions posted only do not.

 

[collinsmark]

16s is rounded to significant figures as my instructor instructed the class on how he wanted it. I personally think I'm losing a bit by doing it but he continually stresses that we do.

Well okay, if that's what your instructor wants. But I'm just saying that 122/8 is actually a little closer to 15 than it is 16.

But apart from the rounding stuff your solution looks okay to me.

I am dumb-founded over this as every example given in class had a stated acceleration, then my homework questions posted only do not.

Oh, but that's part of the joy of physics. By learning some basic principles, you can figure problems out from scratch, even without specific examples. That's one reason why physics is phun!

 

[rwisz]

and stopping a moving object requires a certain amount of time and force. In this situation, your reaction time and the constant acceleration of your car's brakes play a crucial role in determining how long it will take you to stop and how far you will travel before coming to a complete stop. It is understandable that you are having difficulties determining the final time and acceleration in this problem.

One way to approach this problem is to use the equations of motion, specifically the one that relates displacement (x), initial velocity (v1), final velocity (v2), and acceleration (a). This equation is x = (v1 + v2)t/2, where t is the time taken to travel from v1 to v2.

In this case, we know that the initial velocity is 16 m/s and the final velocity is 0 m/s. We also know that the displacement is 130 m. Using this information, we can rearrange the equation to solve for t: t = 2x/(v1 + v2).

Substituting the given values, we get t = 2(130 m)/(16 m/s + 0 m/s) = 16.25 s. This is the total time it takes for you to stop your car at the intersection.

To find the acceleration, we can use the equation a = (v2 - v1)/t. Substituting the values, we get a = (0 m/s - 16 m/s)/16.25 s = -0.98 m/s^2. This means that your car's brakes need to apply a deceleration of 0.98 m/s^2 in order to bring you to a stop at the intersection.

I hope this helps you understand the concept better. It is important to carefully consider all the given variables and use the appropriate equations to solve the problem accurately. Remember, finding time and acceleration in an acceleration problem can be tricky, but with the right approach, you can solve it successfully.