DHS Y4 Math 2 Supplementary Worksheet Integration(Kinematics)

Year 4 Mathematics 2 Applications of Integration - Kinematics Supplementary Worksheet Name : _________________________________ ( ) Class : _______ Date : __________ 1 1 A particle starts from point O and moves in a straight line so that its displacement, s cm, from O, t seconds after leaving O, is given by ( ) 2 6s t t=− . Obtain an expression for the velocity of the particle in terms of t. Hence, determine the value of t when the particle first comes to instantaneous rest and find the acceleration at this instant. The particle is next at O when t = T . Find (i) the value of T, (ii) the distance travelled from t = 0 s to t = T. 2 The height, h m, of a stone t seconds after it has been thrown vertically upwards from ground level is given by 224 3h t t=− . Find (i) its velocity after 3 seconds, (ii) the maximum height reached, (iii) the time of the flight. 3 A particle P starts at a point 3m away from O and travels in a straight line so that its velocity v ms-1 is given by 293v t t=− where t is the time in seconds measured from the start of the motion. Calculate (i) its acceleration when t = 1, (ii) the maximum velocity attained by the particle, (iii) the distance of P from O when it comes to instantaneous rest, (iv) the total distance travelled by P in the first 4 seconds. 4 A particle moves in a straight line so that, at time t seconds after leaving a fixed point O, its velocity , v m s−1, is given by 1 21 2e2 t v − =− . (a) Find (i) the initial acceleration of the particle, (ii) the value of t when the particle is instantaneously at rest, (iii) the distance of the particle from O when t = 2. (b) (i) Sketch the velocity -time curve for t  0, indicating the coordinates of the points of intersection with the axes. (ii) Find the distance travelled during the third second. 5 A particle starts from a point O and moves in a straight line with a velocity v m/s given by sin 2v t t=+ where t seconds is the time after leaving O. (i) Find an expression for the displacement of the particle from O in terms of t, (ii) Calculate the distance travelled by the particle when π 2t = and its acceleration at this instant.

2 6 A particle X moves along a horizontal straight line so that its displacement, s m, from a fixed point O, t seconds after motion has begun, is given by 2328 4 5s t t t= + − − . Obtain expressions, in terms of t, for the velocity and acceleration of X, and state the initial velocity and the initial acceleration of X. A second particle Y moves along the same horizontal straight line as X, and starts from O at the same instant that X begins to move. The initia l velocity of Y is 2 m s -1 and its acceleration, a m s –2 , t seconds after motion has begun, is given by a = 2 – 6t. Find the value of t at the instant when X and Y collide and determine whether or not X and Y are travelling in the same direction at this instant. (Pass GCE ‘O’ Level Examination Additional Mathematics, Shinglee) 7 A particle travels in a straight line with velocity, v m/s, given by v t t= − 1 2 2 where t is the time in seconds after passing a fixed point O. Calculate (i) the distance from O when the acceleration is zero, (ii) the distance travelled by the particle during the 2nd second, (iii) the total distance travelled by the particle after four seconds. 8 The velocity, v m/s, of a particle, t seconds after passing a fixed point O, is given by 2 2 483vt t=− , where 1t . Calculate (i) the acceleration of the particle when 2t= , [2] (ii) the time when it is momentarily at rest, [1] (iii) the total distance moved by the particle for 13 t . [5] [2006 / DHS EOY Y4 P1 / Q17] 9 A particle moves in a straight line so that t seconds after leaving a fixed point O, its velocity, v ms 1− , is given by 2sin 1.vt=− Find (i) the time at which the particle first comes to instantaneous rest, [2] (ii) the distance travelled by the particle in the first 2 seconds. [4] [2008 / DHS EOY Y4 P2 / Q13] Answer 1 v = 3 (t – 6)(t – 2) m/s, t = 2, –12m/s2 (i) 6 (ii) 64 m 2(i) 6 m/s (ii) 48 m (iii) 8 s 3(i) 3 m/s2 (ii) 6.75 m/s (iii) 16.5 m (iv) 19 m 4(a)(i) 1 m/s2 (ii) 2.77 s (iii) 1.53 m (b)(ii) 0.0914 m 5(i) 2 11cos 222s t t= − + (ii) 2.23 m 6 Initial velocity = 4 m/s; Initial acceleration = –10 m/s; 22 3 2Yv t t= − + ; 23 2YS t t t= − + 7 3t = for collision to occur. Both X and Y are travelling in the same direction because both velocities are negative at this instant. 7(i) 1/3 m (ii) 1/3 m (iii) 4 m 8(i) 24 m/s2 (ii) 2 s (iii) 28 m 9(i) π 6 s (ii) 1.09 m