A car describes a distance x metres in time t seconds along a straight road. If the velocity v is constant, then v = ˣ⁄t meter per secThat is, the slope (gradient) of the distance/time graph shown in the figure below is constant.
If, however, the velocity of the car is not constant then the distance/time graph will not be a straight line. It may be as shown in the figure below.
The average velocity over a small time Δt and distance Δx is given by the gradient of the chord AB. That is, the average velocity over time Δt is
As Δt → 0, the chord AB becomes a tangent, such that at point A the velocity is given by v = ᵈˣ⁄dt. Hence the velocity of the car at any instant is given by gradient of the distance/time graph. If an expression for the distance x is known in terms of time, then the velocity is obtained by differentiating the expression.The acceleration a of the car is defined as the rate of change of velocity. A velocity/time graph is shown in the figure below.
If Δv is the change in v and Δt is the corresponding change in time, then
As Δt → 0 the chord CD becomes a tangent such that at the point C, the acceleration is given by a = ᵈⱽ⁄dt.Hence the acceleration of the car at any instant is given by the gradient of the velocity/time graph. If an expression for velocity is known in terms of time t, then the acceleration is obtained by differentiating the expression.Acceleration a = ᵈⱽ⁄dt, where v = ᵈˣ⁄dtHence
The acceleration is given by the second differential coefficient of distance x with respect to time t. The above discussion can be summarised as follows. If a body moves a distance x meters in time t seconds then(i) distance x = f(t).(ii) velocity v = f'(t) or ᵈˣ⁄dt, which is the gradient of the distance/time graph.
Note :(i) Initial velocity means velocity at t = 0.(ii) Initial acceleration means acceleration at t = 0.(iii) If the motion is upward, at the maximum height, the velocity is zero.(iv) If the motion is horizontal, v = 0 when the particle comes to rest.
Solved Problems
Problem 1 :The distance x metres described by a car in time t seconds is given by 😡 = 3t3 – 2t2 + 4t – 1Determine the velocity and acceleration when (i) t = 0 and (ii) t = 1.5 sec.Solution 😡 = 3t3 – 2t2 + 4t – 1Velocity :v = ᵈˣ⁄dtv = 3(3t2) – 2(2t) + 4(1) – 0v = 9t2 – 4t + 4Acceleration :a = ᵈⱽ⁄dta = 9(2t) – 4(1) + 0a = 18t – 4When t = 0,
v = 9(0) – 4(0) + 4v = 4 m/s
a = 18(0) – 4a = m/s2
When t = 1.5 sec.,
v = 9(1.5)2 – 4(1.5) + 4v = 9(2.25) – 6 + 4v = 20.25 – 6 + 4v = 18.25 m/s
a = 18(1.5) – 4a = 27 – 4a = 23 m/s2
Problem 2 :Supplies are dropped from an helicopter and distance fallen in time t seconds is given by xx = (½)gt2where g = 9.8 m/sec2. Determine the velocity and acceleration of the supplies after it has fallen for 2 seconds.Solution :Distance : x = (½)gt2x = (½)(9.8)t2x = 4.9t2Velocity :v = ᵈˣ⁄dtv = 4.9(2t)v = 9.8tAcceleration :a = ᵈⱽ⁄dta = 9.8(1)a = 9.8When t = 2 seconds,
v = 9.8tv = 9.8(2)v = 19.6 m/s
a = 9.8a = 9.8 m/s2
Problem 3 :The angular displacement θ radians of a fly wheel varies with time t seconds and follows the equation θ = 9t2 – 2t3Determine(i) the angular velocity and acceleration of the fly wheel when time t = 1 second and(ii) the time when the angular acceleration is zero.Solution :Part (i) :Angular displacement :θ = 9t2 – 2t3Angular velocity :ω = ᵈθ⁄dtω = 9(2t) – 2(3t2)ω = 18t – 6t2Angular acceleration :α = ᵈω⁄dtα = 18(1) – 6(2t)α = 18 – 12tWhen t = 1 second,
ω = 18(1) – 6(1)2ω = 18 – 6ω = 12 rad/s
α = 18 – 12(1)α = 18 – 12α = 16 rad/s2
Part (ii) :When angular acceleration is zero,α = 018 – 12t = 0-12t = -18t = 1.5 seconds
Problem 4 :A boy, who is standing on a pole of height 14.7 m throws a stone vertically upwards. It moves in a vertical line slightly away from the pole and falls on the ground. Its equation of motion in meters and seconds isx = 9.8t – 4.9t2(i) Find the time taken for upward and downward motions.(ii) Also find the maximum height reached by the stone from the ground.Solution :
Part (i) 😡 = 9.8t – 4.9t2Velocity :v = ᵈˣ⁄dtv = 9.8(1) – 4.9(2t)v = 9.8 – 9.8tAt the maximum height v = 0.9.8 – 9.8t = 0-9.8t = -9.8t = 1 secondSo, the time taken for upward motion is 1 second. For each position x, there corresponds a time t. The ground position is x = -14.7, since the top of the pole is taken as x = 0.To get the total time, substitute x = -14.7 into the given equation.-14.7 = 9.8t – 4.9t2 4.9t2 – 9.8t – 14.7 = 0Multiply both sides by 10. 49t2 – 98t – 147 = 0Divide both sides by 49.t2 – 2t – 3 = 0Factor and solve for t.t2 – 3t + t – 3 = 0t(t – 3) + 1(t – 3) = 0(t – 3)(t + 1) = 0t – 3 = 0 or t + 1 = 0t = 3 or t = -1Time can never be a negative value. So, t = 3.Already we know that the time taken for upward motion is t = 1 second.Therefore, the time taken for downward motion is = 3 – 1 = 2 secondsPart (ii) :When t = 1,x = 9.8(1) – 4.9(1)2x = 9.8 – 4.9x = 9.8The maximum height reached by the stone := 14.7 + 4.9 = 19.6 meters
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