constant acceleration a level maths problem help

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directlink
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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.
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Pangol
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(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.)
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Adh231
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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.
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Pangol
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(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.
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directlink
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(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
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directlink
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(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
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