October 8th, 2016, 02:20 PM  #1 
Member Joined: Mar 2015 From: uk Posts: 33 Thanks: 1  distance speed and acceleration
I'm struggling with this one; can anyone help? A body moves in a straight line with the equation s = t^2 / (1+(t^2)) where t is time in seconds and s is distance travelled in metres. Show that s is greater than or equal to 0 and less than 1. Show also that speed u and acceleration f are given at time t by u = sinx (1+cosx)/2 f = (2cosx  1) (1 + cosx)^2 / 2 where t = tan(x/2) I think I've got the first part by using limits. As t tends to zero then we have 0 / (1+0) = 0 and as t tends to infinity we have (infinity / (1+infinty) which tends to 1. For the second part, speed is rate of change of distance with time or ds/dt. Differentiating using the quotient rule gives ds/dt = 2t / ((1+(t^2))^2) I've then substituted tan(x/2) for t and tried using trig identities to get the answer but without success. Can anyone help with where I'm going wrong? Last edited by skipjack; October 9th, 2016 at 09:48 AM. 
October 8th, 2016, 03:56 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,947 Thanks: 1555 
$u = \dfrac{2t}{(1+t^2)^2}$ sub $t = \tan\left(\dfrac{x}{2}\right)$ ... $u = \dfrac{2\tan\left(\dfrac{x}{2}\right)}{\left[1+\tan^2\left(\dfrac{x}{2}\right)\right]^2}$ $u = \dfrac{2\tan\left(\dfrac{x}{2}\right)}{\sec^4\left (\dfrac{x}{2}\right)}$ $u = 2\tan\left(\dfrac{x}{2}\right) \cos^4\left(\dfrac{x}{2}\right)$ $u = \color{red}{2\sin\left(\dfrac{x}{2}\right) \cos\left(\dfrac{x}{2}\right)} \color{blue}{ \cos^2\left(\dfrac{x}{2}\right)}$ $u = \color{red}{\sin{x}} \cdot \color{blue}{\dfrac{1+\cos{x}}{2}}$ 
October 9th, 2016, 10:10 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,747 Thanks: 2133 
To find f in terms of x, one can use f = du/dt = (du/dx)/(dt/dx).


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acceleration, distance, speed 
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