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Mathematics: Post your doubts here!

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Q.7
(ii)
When t = 2.5
d = 2.6 x 2.5 = 6.5

sin theta = 1.6 / d = 1.6 / 6.5 = 0.246

since a = 10 sin theta
a = 10 x 0.246 = 2.46 m/s'2

could you please explain why you did the a = 10 sin theta bit? thanks :D
 
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why can't we use the polar from to solve it? and also how did you solve for y^2?
we could use the polar form but my answer is not the one given in the markscheme. for y^2, i made a quadratic equation which gave the roots for y^2. To make things easier you could replace the y^2 with any variable such as 'a'
 
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we could use the polar form but my answer is not the one given in the markscheme. for y^2, i made a quadratic equation which gave the roots for y^2. To make things easier you could replace the y^2 with any variable such as 'a'
oh ok thanks alot
 
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dasasas.pngOk if you look in the picture you can see the forces already present

the 4root2 force is the normal to the force applied ON the pulley so if you rearrange the forces then use pythogras to get 2T^2 =32 and solve u get T= 4
 
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