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a) What is the distance it travelled during 2 seconds?
b) In what direction did it ravel (angle with the positive x-axis)?
c) What is the acceleration vector of this particle?
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- Vectors u = −10i + 3j and v = −7i − 9j. What is u − v? a −17i − 6j b 17i + 6j c 3i − 12j d −3i + 12jHere are three vectors in meters: d1 = -6.60î + 8.70ĵ + 3.50k d 2 = - 2.00î – 4.00ĵ + 2.00k d 3 = 2.00î + 3.00ĵ + 1.00k What results from (a) đ1· ( đ 2 + đ 3 ), (b) đ 1 · ( đ 2 × đ 3), and d 1 x ( d 2 + d 3 ) (c), (d) and (e) for i,j and k components respectively)? (a) Number i Units (b) Number Units (c) Number i Units (d) Number Units (e) Number i UnitsYour answer is partially correct. A particle leaves the origin with an initial velocity = (2.54î) m/s and a constant acceleration a ( − 4.67î – 1.95ĵ) m/s². When the particle reaches its maximum x coordinate, what are (a) its velocity, (b) its position vector? (a) Number -1.07 (b) Number Save for Later i î+ Ĵ Units i m/s Ĵ Units
- A particle moves in one dimension according to the equation: x(t)= 2.0m + (5.0m/s)t - (6.2m/s2)t2. Determine x at t=2.0s, vx at t=0.2s, and ax at t=3.1s.A particle is moving in three dimensions and its position vector is given by; r→(t)=(1.6t^2+3.7t)i+(3.7t−1.4)j+(4.6t^3+3.8t)k where r is in meters and ?t is in seconds. Determine the magnitude of the average acceleration between t=0s and t=2s. Express your answer in units of m/s2 using one decimal place.Two vectors are given by đ = 1.3î + 1.8ĵ and b = 7.9î + 1.6j. Find (a) đ x b (b) ở . b ,(c)(a + b and (d) the component of d along the direction of b ? (a) Number i Units (b) Number i Units (c) Number i Units (d) Number Units > > > >
- The equation r(t) = ( sin t)i + ( cos t)j + (t) k is the position of a particle in space at time t. Find the particle's velocity and acceleration vectors. π Then write the particle's velocity at t= as a product of its speed and direction. The velocity vector is v(t) = (i+j+ k.A child starts at one corner of a cubical jungle gym in a playground and climbs up to the diagonally opposite corner. The original corner is the coordinate origin, and the x, y and z axes are oriented along the jungle gym edges. The length of each side is 2 m. The child's displacement is: 2î + 2j + 3.5k 2î+ 2f + 2.8k O 2î + 2j + 2k O 3.5i+ 3.5j+ 3.5k 2.81 + 2.8j + 2kGiven the following vectors: A = -3i; B = i-3j+5k; and C = 7i-15j (i)Find the magnitude and direction of A+C (ii)Find the magnitude and direction of -1.33A + 0.4C. (iii) Find CxB
- From a clifftop over the ocean 130 m above sea level, an object was shot straight up into the air with an initial vertical speed of 166.6 ms On its way down it missed the cliff and fell into the ocean. Its height (above sea level) as time passes can be modeled by the quadratic function f, where f(t)=−4.9t^2+166.6t+130 Here t represents the number of seconds since the object’s release, and f(t) represents the object’s height (above sea level) in meters. 1) After this, this object reached its maximum height. 2) This object flew before it landed in the ocean. 3) This object was above sea level 24s after its release. 4) This object was 1526.5 m above sea level twice: once after its release, and again later after its release.7:31 a 4 O 0令ll回 10. The displacement of the object between t = 0 and t = 4 s is: (A) 22 m (B) 28 m (C) 40 m (D) 42 m (E) 60 m Velocity (mis) Time (b a e 11. The above graph represents the velocity as a function of time of a moving object. What is the net displacement between t = 0 s and t = 4 s? (A) Om (B) 16 m (C) 24 m (D) 36 m (E) 48 m Acceleration vs Time Time (s) a(-5.20)i + (-10.0)j + (18.0)k and then 7.70 s later has V = (-8.20)i + (-10.0)j + (-1.90)k (in meters per second). (a) For that 7.70 s, what is the proton's average acceleration A proton initially has V in unit vector notation, (b) in magnitude, and (c) the angle between a avg and the positive direction of the x axis? avg (a) Number k Units (b) Number Units (c) Number Units