iCoachMath.com: Examples on Position - Velocity - Acceleration Functions - Applications of Derivatives

Curriculum: Alabama Course of Study Click to change Curriculum

Topic: Applications of Derivatives Click to change Topic

Lesson: Position - Velocity - Acceleration Functions Click to change Lesson

Click on a 'View Solution' below for other questions:

e The displacement 'S' described by a particle in t seconds is given by S = 7t - 3t² ft . What is the velocity of the particle at any time t? e

View Solution

e The displacement S of a body in t seconds is given by S = (2t³ - 7t² + 4t + 4) m. The instantaneous velocity of the body is
e

View Solution

e The displacement of a body with time t is related as S = ae^9t - be^-6t. Find the rate of change of displacement with respect to time t. e

View Solution

e The displacement S (in meters) of a particle is related with time t(in seconds) as S = 8t³ + 7t² + 4t + 5 The velocity of the particle when t = 4 sec is e

View Solution

e The displacement S of a particle in time t seconds is given by S = 3t³ - 2t² - 3t + 4 ft . What is the acceleration of the particle at time t? e

View Solution

e What is the acceleration of a body at t = 3 sec whose displacement S(in metres) is related with the time as S = 6t³ - 5t² - 3t + 3? e

View Solution

e The displacement of a particle S (in metres) is related with time t (in seconds) as S = t33 -15t22 + 56t Find the acceleration of the particle when it comes to rest first time.
e

View Solution

e On a straight road, vehicle A started at a point P whose displacement (metres) function with time t (seconds) is S = 2t + 2t². After 1sec vehicle B started from the same point P in the same direction of A. The displacement (metres) function with time t (seconds) of B is S = 4t + 8t². What is the velocity of the vehicle B when it crosses the vehicle A?
e

View Solution

e The displacement function of a particle with time t in seconds is S = 2t³ - 6t² + 2t + 3 ft. Find the velocity of the particle when its acceleration becomes zero.
e

View Solution

e The displacement S of a particle with respect to time t is given by S = - 2t³ + 3t. The velocity of the particle e

View Solution

e The distance 'S' on a straight line traveled by a particle in time t is given by S = 3t - 4t³. What is the maximum velocity of the particle? e

View Solution

e On a straight road, A, B are two mile stones such that AB = a miles. Vehicle P started from A towards B. The displacement (miles) function with time t (hours) of P is S = 5t² + 40t. At the same time vehicle Q started from B towards A. The displacement (miles) function with time t (hours) of Q is s = 5t² + 10t. Find the velocity of P when the vehicles cross each other. [a = 60.]
e

View Solution

e If the square of the displacement S traversed by a particle in time t is given by S² = t² + 7t + 49, then choose the acceleration of the particle from the following.
e

View Solution

e If the displacement function P of a particle with time t is given by P = f(t), then find the acceleration of the particle at any time t.
e

View Solution

e A particle is moving along a straight line according to the law S = 5ut - 2at² where a > 0. Find the greatest and least velocities of the particle. e

View Solution

e The velocity of a particle is related with its displacement S as V² = 2S² + 2kS + k² where k is a constant. What is the acceleration of the particle when it is α units away from the starting point? e

View Solution

e The displacement S of a moving particle is related with time t as S = α t² + β t + c. The displacement of the particle after 1 sec is 40 m, its velocity and acceleration after 2 sec are 90 m/sec and 50 m/sec². Which of the following is correct?
e

View Solution

e The distance S of a body moving along a straight line is related with its velocity v, traversed time t as S = vt + 4. What is the acceleration of the body at any time t in terms of v?
e

View Solution

e If a particle is moving according to the law V² = (5x sin 5x + cos 7x) where V is the velocity and x is the displacement described by it, then what is the acceleration in terms of x? e

View Solution

e The displacement S in metres traversed by a particle with time t in seconds is given by S = 7t² + 8t + 4. Find the average velocity of the particle between t = 0 and t = 4. e

View Solution

e A particle is moving along a straight line so that at time t, its displacement S from a fixed point in the line is given by S = 3α sin(3β t + γ) where α, β, γ are constants. If V and f are the velocity and acceleration of the particle respectively at time t, then v² - fs = ___.
e

View Solution

e The displacement function of a particle with time t is given by S = 3αcos(5βt + γ) where α, β, γ are constants. If V and f are the velocity and acceleration of the particle respectively at time t, then dfdS = ? e

View Solution

e A stone is thrown up vertically and the height S feet reached by it in time t seconds is given by S = 112t - 16t². How much time does the stone take to reach the maximum height?
e

View Solution

e The displacement S covered by a particle in time t is given by S = 4kt + l where k, l are constants, then what is the acceleration of the particle at t = 6?
e

View Solution

e The displacement S of a body in t seconds is given by S = (8t³ - 4t² + 3t + 5) m. The instantaneous velocity of the body is
e

View Solution

e The displacement 'S' described by a particle in t seconds is given by S = 2t - 2t² ft . What is the velocity of the particle at any time t? e

View Solution

e The displacement of a body with time t is related as S = ae^8t - be^-5t. Find the rate of change of displacement with respect to time t. e

View Solution

e The displacement S (in meters) of a particle is related with time t(in seconds) as S = 3t³ + 5t² + 4t + 3 The velocity of the particle when t = 4 sec is e

View Solution

e The displacement S of a particle in time t seconds is given by S = 3t³ - 2t² - 4t + 4 ft . What is the acceleration of the particle at time t? e

View Solution

e What is the acceleration of a body at t = 3 sec whose displacement S(in metres) is related with the time as S = 3t³ - 3t² - 3t + 5? e

View Solution

e The displacement of a particle S (in metres) is related with time t (in seconds) as S = t33 -13t22 + 42t Find the acceleration of the particle when it comes to rest first time.
e

View Solution

e On a straight road, vehicle A started at a point P whose displacement (metres) function with time t (seconds) is S = 3t + 3t². After 1sec vehicle B started from the same point P in the same direction of A. The displacement (metres) function with time t (seconds) of B is S = 6t + 12t². What is the velocity of the vehicle B when it crosses the vehicle A?
e

View Solution

e The displacement function of a particle with time t in seconds is S = 3t³ - 5t² + 5t + 6 ft. Find the velocity of the particle when its acceleration becomes zero.
e

View Solution

e The displacement S of a particle with respect to time t is given by S = - 3t³ + 5t. The velocity of the particle e

View Solution

e The distance 'S' on a straight line traveled by a particle in time t is given by S = 3t - 5t³. What is the maximum velocity of the particle? e

View Solution

e On a straight road, A, B are two mile stones such that AB = a miles. Vehicle P started from A towards B. The displacement (miles) function with time t (hours) of P is S = 3t² + 24t. At the same time vehicle Q started from B towards A. The displacement (miles) function with time t (hours) of Q is s = 3t² + 6t. Find the velocity of P when the vehicles cross each other. [a = 36.]
e

View Solution

e If the square of the displacement S traversed by a particle in time t is given by S² = t² + 9t + 64, then choose the acceleration of the particle from the following.
e

View Solution

e If the displacement function Q of a particle with time t is given by Q = f(t), then find the acceleration of the particle at any time t.
e

View Solution

e A particle is moving along a straight line according to the law S = 2ut - 9at² where a > 0. Find the greatest and least velocities of the particle. e

View Solution

e The velocity of a particle is related with its displacement S as V² = 8S² + 4kS + k² where k is a constant. What is the acceleration of the particle when it is α units away from the starting point? e

View Solution

e The displacement S of a moving particle is related with time t as S = α t² + β t + c. The displacement of the particle after 1 sec is 20 m, its velocity and acceleration after 2 sec are 120 m/sec and 70 m/sec². Which of the following is correct?
e

View Solution

e The distance S of a body moving along a straight line is related with its velocity v, traversed time t as S = vt + 8. What is the acceleration of the body at any time t in terms of v?
e

View Solution

e The displacement S in metres traversed by a particle with time t in seconds is given by S = 2t² + 4t + 5. Find the average velocity of the particle between t = 0 and t = 5. e

View Solution

e A particle is moving along a straight line so that at time t, its displacement S from a fixed point in the line is given by S = 4α sin(5β t + γ) where α, β, γ are constants. If V and f are the velocity and acceleration of the particle respectively at time t, then v² - fs = ___.
e

View Solution

e If a particle is moving according to the law V² = (9x sin 2x + cos 4x) where V is the velocity and x is the displacement described by it, then what is the acceleration in terms of x? e

View Solution

e The displacement function of a particle with time t is given by S = 8αcos(5βt + γ) where α, β, γ are constants. If V and f are the velocity and acceleration of the particle respectively at time t, then dfdS = ? e

View Solution

e A stone is thrown up vertically and the height S feet reached by it in time t seconds is given by S = 72t - 18t². How much time does the stone take to reach the maximum height?
e

View Solution

e The height S feet reached in time t seconds by a stone projected vertically upwards is given by S = 128t - 16t². What is the maximum height reached by the stone?
e

View Solution

e The height S feet reached in time t seconds by a stone projected vertically upwards is given by S = 192t - 16t². What is the maximum height reached by the stone?
e

View Solution