Urooz Ali

A body moves from rest with uniform acceleration and travels 270 metre in 3 seconds | Find the velocity of the body at 10 seconds after the start | Sum Motion in one Dimension – Physics class 9

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Problem : A body moves from rest with uniform acceleration and travels 270 metre in 3 seconds. Find the velocity of the body at 10 seconds after the start. Initial velocity (u)= 0Distance(s) = 270m Time taken (t)=3s final velocity after 10 s (v10)= ? First we will calculate acceleration.We know s = ut + … Read more

When brakes are applied to a bus the retardation produced is 25 cm/s2 and the bus takes 20 s to stop | calculate the initial velocity of bus the distance traveled by bus during this time |

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Problem : When brakes are applied to a bus the retardation produced is 25 centimetre per second square (cm/s2 )and the bus takes 20 seconds to stop. Calculate a)the initial velocity of bus b) the distance traveled by bus during this time Data given:Retardation (-a) = – 25 cm/s2 Time taken to get stopped (tav)=20 … Read more

A car travels a distance 100 metre with a constant acceleration and average velocity of 20 metre per second | The final velocity required by the car is 25 metre per second | Find the initial velocity and acceleration of car | Sum Motion in one Dimension – Physics class 9

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   Problem: A car travels a distance 100 metre with a constant acceleration and average velocity of 20 metre per second. The final velocity required by the car is 25 metre per second. Find 1) the initial velocity and 2) acceleration of car Data given:Distance(s) = 100 metreAcceleration(a) = ? ( to calculate)Average velocity(vav)=20 metre … Read more

Geometry Sum | ABCD is a square. Prove in each case ∆APD ≅ ∆BPC | Find Angle DPC Angle PCD and Angle PCD |

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   ABCD is a square and APB is an equilateral triangle. Prove in each case:  1) &#x2206APD ≅ &#x2206BPC    2) Find the angles of &#x2206DPC Solution : First PartGiven &#x2206ABP = Equilateral triangleR.T.P.1) &#x2206APD ≅ &#x2206BPC    2) Find &#x2220DPC, &#x2220PDC and &#x2220PCD, Proof: In &#x2206APD and &#x2206BPC AP = BP | Same side of … Read more

Theorem | the angle subtended by an arc of a circle | at the centre is double the angle subtended by it at | any point on the remaining part of the circle

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Theorem : A circle  with centre O in which AB subtends  &#x2220AOB at centre and angle &#x2220ACB at any point  on the remaining part of the circle. R.T.P. – &#x2220AOB =  2 x &#x2220ACB Solution : Construction: Join CO and produce CO to point D . 1. In ∆AOC, OC = OA || Radii of … Read more