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- The acceleration vector of a particle is giving by a = -27 t2 j m/s4. The particle is located at the origin at t = 0 s and has an initial velocity Vo = 2(m/s) i + 3(m/s) j. Find: a) The velocity of the particle as a function of time. b) The maximum height the particle reaches.The particle’s position vector varies in time according to the expression ?= 3.00i − 6.00t2? ? (a) Determine the particle’s position and velocity at t = 1.00 s.A particle located initially at (1.5ˆj + 4.0ˆk) m undergoes a displacement of (2.5ˆi+ 3.2ˆj − 1.2ˆk) m. (a) What is the final position of the particle? (b) If the time needed to displace the particle from its initial position to its final position is 5 s, determine the particle’s average velocity vector.
- Initially, a bicycle is at the position 7 = (130 m)i + (210 m)j. After 120 s, the position of the particle is 7 = (110 m)i + (240 m)j . What is the average velocity vector of the bicycle in that time || tinterval? O Tav = (0.17 m/s)i + (-0.25 m/s)j O Jay = (0.25 m/s)i + (-0.17 m/s) O Tay = (-0.25 m/s)î + (0.17 m/s) O Jay = (-0.17 m/s)i + (0.25 m/s)jA projectile is launched from the top of the cliff of height h with initial velocity v0 and launch angle ? with respect to the horizontal. It later hits the ground. Derive an expression for the magnitude and direction angle of the velocity as the object hits the ground in terms of some or all of the variables given in the problem h, v0 , ?, and g.As a projectile moves in its path, is there any point along the path where the velocity and acceleration vectors are parallel to each other? ( assume the object is thrown upward and forward, but not straight up.)
- Suppose the position vector for a particle is given as a function of time by F(c) = x(t)i+y(t)],withx(t)=at+b and y(t)=-ct^2+d, where a=1.20m/s,b=1.10m,c=0.126m/s^2 and d=1.20m. (a) Calculate the average velocity during the time interval from t = 1.90 s to t = 3.85s b) find velocity at t=1.90s Determine the speed at t= 1.90sA particle leaves the origin with an initial velocity v = (3.00i) m/s and a constant acceleration a = (-1.00i – 0.50j) m/s?.When it reaches its maximum x coordinate, what are its (a) velocity and (b) position vector?A projectile is launched with speed v at an angle of ?2 above the horizontal going to a maximum height NH=4.08 H.The projectile is then launched with the same initial speed but an angle of ?1 above the horizontal going to a maximum height H.Please derive the ratio of the total time of flight of the projectile as the time for launch at angle ?2 to that time at angle ?1.
- A particle starts from the origin at t=0, with an initial velocity having an x component of 20m/s and a y component of -15m/s. The particle moves in the xy plane with an x component of acceleration only, given by Ax =4.0 m/s^2 Determine the components of the velocity vector at anytime and the total velocity vector at any time . Calculate the velocity and speed of the particle at t=5.0s.Consider a coordinate system in which the positive x axis is directed upward vertically. What are the positions of a particle (a) 5.0 m directly above the origin and (b) 2.0 m below the origin?A young boy standing next to a lake is throwing rocks at a buoy that is 11 m from the shore. For oneparticular throw, the boy throws a rock from an initial position of 1.0 m above the ground at an angle of 55◦above the horizontal, and after reaching a maximum height of 5.1 m above the water, the rock continues onits parabolic path and lands right on top of the buoy.(a) What is the initial speed of the rock?(b) How tall is the buoy relative to the surface of the water? You can assume that the boy is standing rightat the edge of the water, and the ground where he is standing is level with the surface of the water