Determine the car's displacement

[etSahcs12]

Hi, I am new to the fourm. If it is not any touble, please help me with a kinematic problem.

Homework Statement

A car travels east along a straight road at a constant velocity of 18 m/s. After 5.0s, it accelerates uniformly for 4.0s, reaching a velocity of 24 m/s. For the next 6.0s, the car proceeds with uniform motion. Determine the car's displacement for the 15.0s trip.

Homework Equations

Δd = v1 x Δt + 1/2 x acceleration x Δt(squared)
Δd = v2 x Δt - 1/2 x acceleration x Δt(squared)
Δd = 1/2 (v1 + v2) Δt
2aΔd = v2^2 - v1^2
a = v2 - v2 / Δt

The Attempt at a Solution

This is the only question I have not experienced before. Please guide me through this question.

 

[Kurdt]

You will have to split the problem into 3 parts. There are two where the car is traveling with a constant velocity and one where it is accelerating. The equation you have stated is almost correct. For constant velocity you will need:

d = vt

For constant acceleration you will need one of the constant acceleration kinematic equations.

https://www.physicsforums.com/showpost.php?p=905663&postcount=2

 

[etSahcs12]

so do i do:

18 m/s x 5.0s then 24 x 4.0s then add them up?

What do i do with the 6.0s and the uniform motion? When adding do i find the displacement?

 

[Kurdt]

so do i do:

18 m/s x 5.0s then 24 x 4.0s then add them up?

What do i do with the 6.0s and the uniform motion? When adding do i find the displacement?

No. The two time intervals when the car is traveling with constant velocity are the first 5 seconds and the last 6 seconds after the acceleration. To work out the distance traveled while accelerating you will need one of the kinematic equations in the link I gave you above. To save you the time here it is:

d = ut +\frac{1}{2} at^2

 

[etSahcs12]

How am i able to find the acceleration and v1 by using all of these variables (specifically time and velocity)? Would i use d = vt at all?

 

[HallsofIvy]

am i supposed to divide this question into 3 parts as well?

Which question are you talking about? You have posted on question and were told to break it into 3 parts.

A car travels east along a straight road at a constant velocity of 18 m/s. After 5.0s,

Okay, how far will it go at 18 m/s for 5 s?

it accelerates uniformly for 4.0s, reaching a velocity of 24 m/s.

Hint: at uniform acceleration, the "average speed" is just the average of the lowest and highest speeds: (18+24)/2 m/s. How far will the car go in 4.0s at that average speed?

For the next 6.0s, the car proceeds with uniform motion. Determine the car's displacement for the 15.0s trip.

At 24 m/s how far will the car go in 6.0 s? NOW add them all up.

 

[etSahcs12]

Ohh i see. If i were to add them, I would get the displacement correct? or should i use d = ut +\frac{1}{2} at^2 after adding.

 

[Kurdt]

or should i use d = ut +\frac{1}{2} at^2 after adding.

You can ignore that one, I made a mistake. You will need to use the average velocity as HallsofIvy suggested in d = vt.

 

[etSahcs12]

so, bastically i do d = vt with the time and velocities, add them up and that's my displacement?

 

[fanj06]

Please help me with this proble. I have tried everything.
A rifle is aimed horizontally at a target located 44.0 m away. The bullet hits the target 3.0 cm below the aim point. What was the bullet's time of flight?

 

[Kurdt]

so, bastically i do d = vt with the time and velocities, add them up and that's my displacement?

Yes but remember the middle time stint will require you work out the average speed.

Please help me with this proble. I have tried everything.
A rifle is aimed horizontally at a target located 44.0 m away. The bullet hits the target 3.0 cm below the aim point. What was the bullet's time of flight?

Please do not hijack other peoples threads. You can start a new thread of your own if you have a question.

 

[etSahcs12]

I see, so after adding 24+18 and dividing by 2, multply the time and that will be my distance ? Once done, do the same for the other 2 (but without finding the average acceleration) and add, thus giving my displacement?

 

[Kurdt]

I see, so after adding 24+18 and dividing by 2, multply the time and that will be my distance ? Once done, do the same for the other 2 (but without finding the average acceleration) and add, thus giving my displacement?

Yes.