# A Level Mechanics Kinematics question

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What Just a question regarding Kinematics in A Level Mechanics.

Say a ‘particle’ was thrown upwards off the edge of a cliff 100m high at 8ms^-1, and say the time of the journey was meant to be calculated (with the effect air resistance ignored). In M1 I was taught to using the Kinematics formula twice for each part of the journey, on the way up to the maximum and then again for its journey down to the plane.

For its journey to the maximum:

u = 8

a = -9.8 (deceleration due to gravity)

v = 0

t = ?

Using v^2 = u^2 + 2as:

s = 3.27m (distance to maximum)

Using v = u + at:

t = 0.82s (time to maximum)

Then on the journey from the maximum to the plane:

u = 0

a = +9.8

s = 103.27

Using s = ut + 0.5at^2:

t = 4.59s (time from maximum to plane)

Therefore the total time of the journey would be 5.41s (4.59 + 0.82)

However, in M2 (particularly with Projectile Motion) we’ve been told we can take the acceleration as negative for the whole journey.

In the same example:

u = 8

a = -9.8

s = -100

Using s = ut + 0.5at^2:

t = 5.41s

This proves that calculating the time of the journey for each part of it (from top off cliff to maximum, then from maximum to plane) gives the same answer as calculating the time of the whole journey in one go.

However, I don’t understand the logic behind this, as it includes taking the acceleration as a negative constant for the whole journey (as its value changes from negative or positive past the maximum).

Say a ‘particle’ was thrown upwards off the edge of a cliff 100m high at 8ms^-1, and say the time of the journey was meant to be calculated (with the effect air resistance ignored). In M1 I was taught to using the Kinematics formula twice for each part of the journey, on the way up to the maximum and then again for its journey down to the plane.

For its journey to the maximum:

u = 8

a = -9.8 (deceleration due to gravity)

v = 0

t = ?

Using v^2 = u^2 + 2as:

s = 3.27m (distance to maximum)

Using v = u + at:

t = 0.82s (time to maximum)

Then on the journey from the maximum to the plane:

u = 0

a = +9.8

s = 103.27

Using s = ut + 0.5at^2:

t = 4.59s (time from maximum to plane)

Therefore the total time of the journey would be 5.41s (4.59 + 0.82)

However, in M2 (particularly with Projectile Motion) we’ve been told we can take the acceleration as negative for the whole journey.

In the same example:

u = 8

a = -9.8

s = -100

Using s = ut + 0.5at^2:

t = 5.41s

This proves that calculating the time of the journey for each part of it (from top off cliff to maximum, then from maximum to plane) gives the same answer as calculating the time of the whole journey in one go.

However, I don’t understand the logic behind this, as it includes taking the acceleration as a negative constant for the whole journey (as its value changes from negative or positive past the maximum).

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What Just a question regarding Kinematics in A Level Mechanics.

Say a ‘particle’ was thrown upwards off the edge of a cliff 100m high at 8ms^-1, and say the time of the journey was meant to be calculated (with the effect air resistance ignored). In M1 I was taught to using the Kinematics formula twice for each part of the journey, on the way up to the maximum and then again for its journey down to the plane.

For its journey to the maximum:

u = 8

a = -9.8 (deceleration due to gravity)

v = 0

t = ?

Using v^2 = u^2 + 2as:

s = 3.27m (distance to maximum)

Using v = u + at:

t = 0.82s (time to maximum)

Then on the journey from the maximum to the plane:

u = 0

a = +9.8

s = 103.27

Using s = ut + 0.5at^2:

t = 4.59s (time from maximum to plane)

Therefore the total time of the journey would be 5.41s (4.59 + 0.82)

However, in M2 (particularly with Projectile Motion) we’ve been told we can take the acceleration as negative for the whole journey.

In the same example:

u = 8

a = -9.8

s = -100

Using s = ut + 0.5at^2:

t = 5.41s

This proves that calculating the time of the journey for each part of it (from top off cliff to maximum, then from maximum to plane) gives the same answer as calculating the time of the whole journey in one go.

However, I don’t understand the logic behind this, as it includes taking the acceleration as a negative constant for the whole journey (as its value changes from negative or positive past the maximum).

**Silxnt1**)What Just a question regarding Kinematics in A Level Mechanics.

Say a ‘particle’ was thrown upwards off the edge of a cliff 100m high at 8ms^-1, and say the time of the journey was meant to be calculated (with the effect air resistance ignored). In M1 I was taught to using the Kinematics formula twice for each part of the journey, on the way up to the maximum and then again for its journey down to the plane.

For its journey to the maximum:

u = 8

a = -9.8 (deceleration due to gravity)

v = 0

t = ?

Using v^2 = u^2 + 2as:

s = 3.27m (distance to maximum)

Using v = u + at:

t = 0.82s (time to maximum)

Then on the journey from the maximum to the plane:

u = 0

a = +9.8

s = 103.27

Using s = ut + 0.5at^2:

t = 4.59s (time from maximum to plane)

Therefore the total time of the journey would be 5.41s (4.59 + 0.82)

However, in M2 (particularly with Projectile Motion) we’ve been told we can take the acceleration as negative for the whole journey.

In the same example:

u = 8

a = -9.8

s = -100

Using s = ut + 0.5at^2:

t = 5.41s

This proves that calculating the time of the journey for each part of it (from top off cliff to maximum, then from maximum to plane) gives the same answer as calculating the time of the whole journey in one go.

However, I don’t understand the logic behind this, as it includes taking the acceleration as a negative constant for the whole journey (as its value changes from negative or positive past the maximum).

In your first approach, you split the journey into two parts. In the first part, UPWARDS was your +ve direction hence you chose a = -9.8 because, as I said, acceleration acts down hence the minus since it's going into the opposite direction from the one you assume was +ve. In the second part, DOWNWARDS was your +ve direction hence you chose a=9.8 as required as now its direction agrees with your +ve direction.

You don't *have* to change your sense of +ve direction when it's at a maximum, and in fact I wouldn't suggest doing so because you lose consistency in your method hence your chance of making errors is greater. The point of this split and change is that some people find it easier to think about this "when the particle is moving up, I'll just take UPWARDS to be +ve, but when it's moving down I'll take DOWNWARDS as +ve" but it's really unnecessary mathematically.

In your second approach, you just stick with the agreement that UPWARDS is the +ve direction throughout the entire journey hence a=-9.8 throughout it all, and it's also why you say s=-100 as well, note how this is now negative too because the bottom of the cliff is downwards from the starting position which is measured in the opposite direction from your +ve UPWARDS direction. In order to find the time when the particle is at its max, you can still use SUVAT but just realise that the velocity of the particle is zero at the maximum point hence just use v=u+at and solve 0=8-9.8t for the same time you got in your first approach.

Last edited by RDKGames; 1 year ago

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(Original post by

Acceleration due to gravity always acts down, and just like with any other question of this type you need to decide which direction you will take to be your +ve direction.

In your first approach, you split the journey into two parts. In the first part, UPWARDS was your +ve direction hence you chose a = -9.8 because, as I said, acceleration acts down hence the minus since it's going into the opposite direction from the one you assume was +ve. In the second part, DOWNWARDS was your +ve direction hence you chose a=9.8 as required as now its direction agrees with your +ve direction.

You don't *have* to change your sense of +ve direction when it's at a maximum, and in fact I wouldn't suggest doing so because you lose consistency in your method hence your chance of making errors is greater. The point of this split and change is that some people find it easier to think about this "when the particle is moving up, I'll just take UPWARDS to be +ve, but when it's moving down I'll take DOWNWARDS as +ve" but it's really unnecessary mathematically.

In your second approach, you just stick with the agreement that UPWARDS is the +ve direction throughout the entire journey hence a=-9.8 throughout it all, and it's also why you say s=-100 as well, note how this is now negative too because the bottom of the cliff is downwards from the starting position which is measured in the opposite direction from your +ve UPWARDS direction. In order to find the time when the particle is at its max, you can still use SUVAT but just realise that the velocity of the particle is zero at the maximum point hence just use v=u+at and solve 0=8-9.8t for the same time you got in your first approach.

**RDKGames**)Acceleration due to gravity always acts down, and just like with any other question of this type you need to decide which direction you will take to be your +ve direction.

In your first approach, you split the journey into two parts. In the first part, UPWARDS was your +ve direction hence you chose a = -9.8 because, as I said, acceleration acts down hence the minus since it's going into the opposite direction from the one you assume was +ve. In the second part, DOWNWARDS was your +ve direction hence you chose a=9.8 as required as now its direction agrees with your +ve direction.

You don't *have* to change your sense of +ve direction when it's at a maximum, and in fact I wouldn't suggest doing so because you lose consistency in your method hence your chance of making errors is greater. The point of this split and change is that some people find it easier to think about this "when the particle is moving up, I'll just take UPWARDS to be +ve, but when it's moving down I'll take DOWNWARDS as +ve" but it's really unnecessary mathematically.

In your second approach, you just stick with the agreement that UPWARDS is the +ve direction throughout the entire journey hence a=-9.8 throughout it all, and it's also why you say s=-100 as well, note how this is now negative too because the bottom of the cliff is downwards from the starting position which is measured in the opposite direction from your +ve UPWARDS direction. In order to find the time when the particle is at its max, you can still use SUVAT but just realise that the velocity of the particle is zero at the maximum point hence just use v=u+at and solve 0=8-9.8t for the same time you got in your first approach.

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