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Mathematics: Post your doubts here!

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For
dhwyFt6.jpg

Variable Acceleration
 
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I dont understand, how did you get that? :s
component is mgsin(theta)
mg is 12500
For sintheta, consider the triangle in the question. Length of the slope(500) is hypotenuse and the height(30) is the perpendicular. sin(theta) = 30/500
so the component is 12500(30/500)
 
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component is mgsin(theta)
mg is 12500
For sintheta, consider the triangle in the question. Length of the slope(500) is hypotenuse and the height(30) is the perpendicular. sin(theta) = 30/500
so the component is 12500(30/500)
Ohhh, got it! thankyou so much!
 
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Velocity after A is 10t - 0.15t^2 +C (Integrated acceleration)
This will be equal to 5 as at t=0 P is at A and that instant its velocity is 5ms-1.
So velocity after A is 10t - 0.5t^2 +5
Integrate this to get distance
5t^2 - 0.05t^3 + 5t
Time taken from O to A was 0.5s so after A will be 3.0s.
Put 2.5 in the equation to get 42.97 m after A.
Add 1.25 m from O to A. So total distance is 44.2 m.
 
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Velocity after A is 10t - 0.15t^2 +C (Integrated acceleration)
This will be equal to 5 as at t=0 P is at A and that instant its velocity is 5ms-1.
So velocity after A is 10t - 0.5t^2 +5
Integrate this to get distance
5t^2 - 0.05t^3 + 5t
Time taken from O to A was 0.5s so after A will be 3.0s.
Put 2.5 in the equation to get 42.97 m after A.
Add 1.25 m from O to A. So total distance is 44.2 m.
Oh ok thanks! i was taking time as 0.5s instead of 0 :S
 
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