# constant acceleration a level maths problem help

Watch
Announcements
Thread starter 1 year ago
#1
a van passes a car. the van is travelling 27m/s and the car is travelling 15m/s. from the instant the van is level with the car, the car accelerates uniformly at 3m/s/s in order to catch up to the van. Work out the time taken until the car is level with the van.
0
reply
1 year ago
#2
(Original post by directlink)
a van passes a car. the van is travelling 27m/s and the car is travelling 15m/s. from the instant the van is level with the car, the car accelerates uniformly at 3m/s/s in order to catch up to the van. Work out the time taken until the car is level with the van.
Since the van and the car are level when the problem starts, they will have travelled the same distance at the point where the car catches up with the van. Can you find an expression for the distance travelled by the car and the distance travelled by the van at any time after they are first level? If so, you can put these expressions equal to each other. (Note that you do not have to find a value for this distance.)
0
reply
1 year ago
#3
Ignore everything before the car and van are together.

Consider the car,
s=s, u=15, a=3, t=t.
With this information and using SUVAT you can form the equation: s=15t+1.5t^2
Consider the van,
s=s, u=27, a=0, t=t
Again you form another equation, this time s=27t

As you know both the car and van must have the same displacement from the first meeting point to the second meeting point you can equate the equations
27t=15t+1.5t^2.
This equation you can then solve, to work out the value of t.
0
reply
1 year ago
#4
(Original post by Adh231)
Ignore everything before the car and van are together.

Consider the car,
s=s, u=15, a=3, t=t.
With this information and using SUVAT you can form the equation: s=15t+1.5t^2
Consider the van,
s=s, u=27, a=0, t=t
Again you form another equation, this time s=27t

As you know both the car and van must have the same displacement from the first meeting point to the second meeting point you can equate the equations
27t=15t+1.5t^2.
This equation you can then solve, to work out the value of t.
It's better to give hints than (almost) fully worked solutions.
0
reply
Thread starter 1 year ago
#5
(Original post by Pangol)
Since the van and the car are level when the problem starts, they will have travelled the same distance at the point where the car catches up with the van. Can you find an expression for the distance travelled by the car and the distance travelled by the van at any time after they are first level? If so, you can put these expressions equal to each other. (Note that you do not have to find a value for this distance.)
uh would it be 27t for the van and 15t for the car
0
reply
Thread starter 1 year ago
#6
(Original post by Adh231)
Ignore everything before the car and van are together.

Consider the car,
s=s, u=15, a=3, t=t.
With this information and using SUVAT you can form the equation: s=15t+1.5t^2
Consider the van,
s=s, u=27, a=0, t=t
Again you form another equation, this time s=27t

As you know both the car and van must have the same displacement from the first meeting point to the second meeting point you can equate the equations
27t=15t+1.5t^2.
This equation you can then solve, to work out the value of t.
i appreciate u thank u
0
reply
X

Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Have you experienced financial difficulties as a student due to Covid-19?

Yes, I have really struggled financially (29)
15.18%
I have experienced some financial difficulties (57)
29.84%
I haven't experienced any financial difficulties and things have stayed the same (75)
39.27%
I have had better financial opportunities as a result of the pandemic (25)
13.09%
I've had another experience (let us know in the thread!) (5)
2.62%

View All
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.