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The velocity V m/s of a particle in motion is given by V=3t2-2t+5.Calculate the distance travelled by the particle between t= 2 seconds and t = 6 seconds. (3 marks)
Form 4 Mathematics
Find the coordinates of the turning point of the curve y= x²-14x +10
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Form 4 Mathematics
The equation of a curve is y=x3+x2-x-1
(i) Determine the stationary point of the curve (Îi) the nature of the stationary points in (a) (i) above. (b) Determine: (i) the equation of the tangent to the curve at x = 1; (ii) the equation of the normal to the curve at x = 1.
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Form 4 MathematicsA curve is represented by the function y = 1/3 x3 + x2 – 3x + 2 (a) Find dy/dx (1 mark) (b) Determine the values of y at the turning points of the curve y = 1/3 x3 + x2 – 3x + 2 ( 4 marks)
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Form 4 Mathematics
The equation of a curve is given as y=1/3x3-4x+5
Determine: (a) The value of y when x = 3; (b) The gradient of the curve at x = 3; (c) The turning points of the curve and their nature.
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Form 4 Mathematics
The distance covered by a moving particle through point O is given by the equation, s = t3 - 15t2 + 63t — 10.
Find: (a) distance covered when t = 2 (b) the distance covered during the 3rd second; (c) the time when the particle is momentarily at rest; (d) the acceleration when t = 5.
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Form 4 MathematicsA stone is thrown vertically upwards from a point O After t seconds, the stone is S metres from O Given that S= 29.4t – 4.9t2, Find the maximum height reached by the stone ( 3 marks)
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Form 4 MathematicsThe gradient of a curve at point (x,y) is 4x – 3. the curve has a minimum value of – 1/8 (a) Find (i) The value of x at the minimum point ( 1 mark) (ii) The equation of the curve ( 4 marks) (b) P is a point on the curve in part (a) (ii) above. If the gradient of the curve at P is -7, find the coordinates of P ( 3 marks)
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Form 4 Mathematics
The equation of a curve is given as y = 2x3 -9/2 x2 -15x + 3.
(a) Find: (i) the value of y when x = 2; (ii) the equation of the tangent to the curve at x = 2. (b) Determine the turning points of the curve.
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Form 4 Mathematics
A particle moves in a straight line. It passes though point O at t = O with velocity v= 5m/s. The acceleration a m/s2 of the particle at time t seconds after passing through O is given by a = 6t + 4
(a) Express the velocity v of the particle at time t seconds in terms of t (b) Calculate (i) The velocity of the particle when t = 3 (ii) The distance covered by the particle between t = 2 and t = 4
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Form 4 Mathematics
The equation of a curve is y = 2x3 + 3x2.
a)Find i)The x – intercept of the curve ii) they-intercept of the curve b i)Determine the stationary points of the curve ii)For each point in (b) (i) above, determine whether it is a maximum or a minimum c) Sketch the curve.
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Form 4 Mathematics
The gradient of the tangent to the curve y = ax3 + bx at the point (1,1) is -5 , Calculate the values of a and b.
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Form 4 Mathematics
A particle was moving along a straight line. The acceleration of the particle after t seconds was given by (9 -3t) ms-2. The initial velocity of the particle was 7 ms-1.
Find: a) the velocity (v) of the particle at any given time (t); b) The maximum velocity of the particle; c)the distance covered by the particle by the time it attained maximum velocity
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Form 4 Mathematics
The gradient of the curvey y = 2x3 – 9x2 + px – 1 at x = 4 is 36.
a)Find : i) the value of p; ii)The equation of the tangent to the curve at x = 0.5. b) Find the coordinates of the training points of the curve
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Form 4 Mathematics
The velocity V ms, of a moving body at time t seconds is given by V = 5t2 – 12t + 7
Find its acceleration after 2 seconds.
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Form 4 Mathematics
The gradient of a curve is given by dy/dx = x2 - 4x4- 3. The curve passes through the point (1,0). Find the equation of the curve.
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Form 4 Mathematics
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Form 4 Mathematics
The equation of acurve is given by y = x3 – 4x2 – 3x
(a) Find the value of y when x = -1 (b) Determine the stationary points of the curve (c) Find the equation of the normal to the curve at x = 1
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Form 4 MathematicsThe velocity Vms-1 of particle in motion is given by V =3t2 – t +4, where t is time in seconds. Calculate the distance traveled by the particle between the time t=1 second and t=5 seconds.
aNSWER
Form 4 Mathematics
The displacement, s metres, of a moving particle from a point O, after t seconds is given by, s = t3 – 5t2 + 3t + 10
a) Find s when t =2. b) Determine: i. The velocity of the particle when t = 5 seconds; ii. The value of t when the particle is momentarily at rest. c) Find the time, when the velocity of the particles is maximum.
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Form 4 Mathematics
The acceleration of a body moving along a straight line is (4 - t) m/s2 and its velocity is v m/s after t seconds.
(a) (i) If the initial velocity of the body is 3 m/s, express the velocity v in terms of t. (ii) Find the velocity of the body after 2 seconds. (b) Calculate: (i) the time taken to attain maximum velocity; (ii) the distance covered by the body to attain the maximum velocity.
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Form 4 Mathematics
The displacement s metre of a particle moving along straight line after t seconds is given by. S = 3t + 3/2 t2 – 2t3
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