Warning: Illegal string offset 'html' in /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php on line 909

Distance, Speed and Time Maybe - HSN forum

Distance, Speed and Time Maybe

6 replies to this topic

#1Karennm

Showing Improvement

• Members
• 27 posts
• Location:Kilmacolm, Scotland
• Gender:Female

Posted 18 April 2009 - 10:51 AM

A ball is thrown vertically upwards.
After t seconds its height is h metres, where h=1.2+19.6-4,9t^2

(a) Find the speed of the ball after 1 second.
(b) For how many seconds is the ball travelling upwards?

#2Cognitive

Newbie

• Members
• 8 posts
• Location:Scotland
• Interests:Stuff :-)
• Gender:Not Telling

Posted 18 April 2009 - 02:12 PM

I think you may have made a mistake in typing out the equation, should it be:

h = 1.2 + 19.6t - 4.9t^2 ???

Assuming it is the above:

a) You need to find an equation which describes the speed of the object. You need to know that velocity (speed) is the first derivative with respect to time of displacement (height in this case).

So basically, you need to find the derivative of the equation h. You then need to sub in 1 and evaluate it.

b) As the height of the ball is modelled by a quadratic equation, and the coefficient infront of the squared term is negative, a MAXIMUM exists. This is the maximum height. When the ball is at this maximum, its velocity will be 0. You should equate your velocity function with 0, and then solve to find t. The time you get will be the time at which the ball is at its maximum height. This is equivalent to the time it has spent travelling upwards

Just incase this ever comes up in another question, acceleration is the derivative of velocity with respect to time (basically the second derivative of displacement with respect to time)

Hope that's all correct and that it helps.

#3Karennm

Showing Improvement

• Members
• 27 posts
• Location:Kilmacolm, Scotland
• Gender:Female

Posted 19 April 2009 - 10:52 AM

QUOTE (Cognitive @ Apr 18 2009, 03:12 PM) <{POST_SNAPBACK}>
I think you may have made a mistake in typing out the equation, should it be:

h = 1.2 + 19.6t - 4.9t^2 ???

Assuming it is the above:

a) You need to find an equation which describes the speed of the object. You need to know that velocity (speed) is the first derivative with respect to time of displacement (height in this case).

So basically, you need to find the derivative of the equation h. You then need to sub in 1 and evaluate it.

b) As the height of the ball is modelled by a quadratic equation, and the coefficient infront of the squared term is negative, a MAXIMUM exists. This is the maximum height. When the ball is at this maximum, its velocity will be 0. You should equate your velocity function with 0, and then solve to find t. The time you get will be the time at which the ball is at its maximum height. This is equivalent to the time it has spent travelling upwards

Just incase this ever comes up in another question, acceleration is the derivative of velocity with respect to time (basically the second derivative of displacement with respect to time)

Hope that's all correct and that it helps.

Im having problem factorising it?

#4Cognitive

Newbie

• Members
• 8 posts
• Location:Scotland
• Interests:Stuff :-)
• Gender:Not Telling

Posted 19 April 2009 - 03:18 PM

You don't need to factorise anything.

When you differentiate the height equation, you get an equation for velocity (v), so:

v = 19.6 - 9.8t

To find the time when it's at its maximum, you set v = 0, so:

19.6 -9.8t = 0
9.8t = 19.6
t = 19.6/9.8 = 2 seconds

Hope that helps

#5Karennm

Showing Improvement

• Members
• 27 posts
• Location:Kilmacolm, Scotland
• Gender:Female

Posted 19 April 2009 - 04:12 PM

Aww Yeh Thats Good Thanks

#6davidmarsh01

Newbie

• Members
• 8 posts
• Gender:Male

Posted 01 September 2009 - 04:23 PM

is this not more of a physics question?

#7Nathan

Fully Fledged Genius

• Members
• 1,736 posts
• Location:Aberdeen, Scotland
• Gender:Male

Posted 03 September 2009 - 08:01 AM

It is, but it's also covered in the higher maths syllabus

1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users