Deceleration Problem: Solving for Time, Distance, and Trajectory

[cellfinder1]

I have a question that I have been working on, but I am stuck and not quite sure if i am approaching the problem right.
The question asks: a person is driving 30.0 m/s 23 degrees north of east when a person walks out in front of them. The person slams their brakes and turns the cars wheels. The result is a deceleration(constant) that is 5.00 m/s2 oriented 53 degree north of east. How long does it take to stop? B)How far does the car go before stopping? C)What is the car trajectory?

What I first did was draw it up,and assumed to take the 30 m/s 23 degrees north of east and finding the components. Vx and Vy. They were Vx=27.6m/s and Vy=11.7 m/s. Then itried to find the time, hoping too find the distance. But the whole thing is I do not know what to do with the deceleration part and the 53 degrees that was given in the question. Any help would be helpful!

 

[AKG]

At first glance, I would say that rather than resolving it into Vx and Vy, resolve it into its velocity in the direction of acceleration (53 degrees north of east) and it's velocity perpendicular to that direction. You could forget about direction and treat the things as scalars, where in one direction, the car would be undergoing constant acceleration and in the other, constant velocity. However, noting that second bit, that in one direction it will be undergoing constant velocity, I had to look over the problem again. What I think they're saying is that as soon as the driver noticed the person, he started heading 53 degrees north of east, i.e. the same speed, 30.0 m/s, but a different direction. So, at first he's going at a constant velocity in one direction, and then he suddenly starts heading in another direction but this time at constant acceleration (and obviously decreasing speed). So, what I just said basically describes, in words, the car's trajectory. As for A and B, you can just use your basic kinematic equations, and again, you can treat the stuff in scalars because the velocity and acceleration would be colinear. You know initial speed/velocity (30.0), final velocity (0), and acceleration (-5.00). All you have to do is calculate time elapsed and distance travelled, and I'm sure you know the equations to solve those.

 

[M98Ranger]

Hello there,

Thank you for reaching out for help with your deceleration problem. It seems like you have made some progress in breaking down the given information and finding the components of the initial velocity. However, you are correct in feeling stuck when it comes to the deceleration and the 53 degrees given in the question.

To solve this problem, you will need to use the equations of motion for constant acceleration. These equations are:

1) v = u + at
2) s = ut + 1/2 at^2
3) v^2 = u^2 + 2as

Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
s = distance

Since you are given the initial velocity, acceleration, and the angle at which the car is decelerating, you can use the components of the initial velocity (Vx and Vy) to find the acceleration in the x and y directions. This can be done using basic trigonometry.

Once you have the acceleration in the x and y directions, you can use it in the above equations to solve for the unknowns.

For part A) How long does it take to stop?
You can use equation 1) to solve for t. Remember to use the component of the acceleration in the direction of the initial velocity.

For part B) How far does the car go before stopping?
You can use equation 2) to solve for s. Again, use the component of the acceleration in the direction of the initial velocity.

For part C) What is the car trajectory?
To find the car trajectory, you will need to use the components of the initial velocity and the acceleration in the x and y directions to find the final velocity in the x and y directions. Then, you can use basic trigonometry to find the final velocity and the angle at which the car is moving after decelerating.

I hope this helps you in approaching the problem and finding the solution. Remember to always break down the given information and use the appropriate equations to solve for the unknowns. Good luck!