# Motion under uniform acceleration for a particle

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MOTION UNDER UNIFORM ACCELERATION For a particle moving under uniform acceleration, the following equations apply; 1. v = u +at 2. V 2 = u 2 + 2as 3. s = ut + at 2 4. s = ௨ା௩ 2 Where u = Initial velocity v = final velocity, a = acceleration, s = displacement, t = time taken. EXAMPLES 1.A car starting from rest is accelerated at the rate of 0.4m/s 2 . Find the distanced covered by the car in 10 seconds? Before any calculation is done it is important to extract data from the the given question.
Prepared By: ChilesheMwenyaMporokoso Page 9 In starting from rest it means u = 0m/s. So have U = 0m/s, a = 0.4m/s 2 , t = 10s,S = ? Using the equation S = ut + at 2 = 0x10 + x0.4x (10) 2 = 20m 2. A motor bike initially travelling with a velocity of 40m/s is slowed down by applying brakes. If the retardation is 8m/s 2 , Determine; (i) The velocity after covering 50m, (ii) How long it takes to cover 50m. SOLUTION u = 40m/s, retardation is negative acceleration. a = -8m/s 2 s = 50m using the equation V 2 = u 2 + 2as V 2 = (40) 2 +2 x (50) 2 = 800 V = √800 = 28.3m/s From the equation V = u +at, we make t the subject of the formula. t = ௩ି௨ = ଶ଼.ଷିସ଴ ି଼ t = 1.46seconds. Graphs can be used to represent motion of objects moving in a straight line. And problems can be solved using graphs.
Prepared By: ChilesheMwenyaMporokoso Page 10 Examples 1.A train accelerates uniformly from rest to reach a velocity of 54km/h in 200 Seconds, after which the velocity remained constant for 300 seconds. At the End of this time, the train decelerates to rest in 150 seconds. Sketch the velocity – time graph and find the total distance travelled. SOLUTION We can tabulate the data given in the question. 54km/h = 54 x ଵ଼ = 15m/s Velocity 0 15 15 0 Time 0 200 500 650 The sketch of the tabulated data is as shown below. For freely falling objects, the same equations of motion apply except: a g s h Examples (i)A ball is thrown vertically upwards at 40m/s,calculate:
Prepared By: ChilesheMwenyaMporokoso Page 11 a.)The greatest height reached b.)The time taken to return to the ground. Solution Data (a)u = 40m/s v = 0m/s. Anything thrown upwards has a velocity equal to zero at maximum Height g = - 9.81m/s 2 . Using the equation v 2 = u 2 + 2gh h = ି ୳ ଶ୥ = ሺ଴ሻ ି ሺସ଴ሻ ଶ ୶ିଽ.଼ଵ = 81.55m (b)Using the equation V = u +at and making t the subject of the formula, we have T = ௩ି௨ = ଴ିସ଴ ିଽ.଼ଵ = 4.08sec.This the time taken to go up The time taken to go up is equal to time taken to come down Thus total time taken for all journey t = 4.08x 2 = 8.16sec
Prepared By: ChilesheMwenyaMporokoso Page 12 (ii)A stone is thrown upwards with a velocity of 4.9m/s from the top of the bridge. If it falls down in water after 2 seconds, find the height of the bridge. Solution Data u = 4.9m/s v= 0m/s g = -9.81 t = 2sec Using the equation h = ut + gt 2 = 4.9 x 2 + x (-9.81) x (2) 2 = -9.82m The height of the bridge is 9.82m The negative sign is an indication that the height is below the throwing point
Prepared By: ChilesheMwenyaMporokoso Page 13 Self evaluation exercise 1. A car starts from rest and attains a speed of 40m/s in 20 seconds.

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