please i need help asap Problem 1 The acceleration of a particle moving only on

Question

please i need help asap

Problem 1

The acceleration of a particle moving only on a horizontal xy plane is given by a=3ti+4tj, where a is in meters per seconds squared and t is in seconds, at t=0, the position vector r=(20.0m)i+(40.0m)j locates the particles, which then has the velocity vector v=(5.00m/s)i+(2.00m's)j. at t=4.00s, what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the positive direction of the x axis?

Problem 2

A particle starts from the origine at t=0 with a velocity of 8.0j m/s and moves in the xy plane with constant acceleration (4.0i+2.0j)m/s^2. when the particle's x coordinate is 29m, what are its (a) y coordinate and (b) speed?

Explanation / Answer

a=3ti+4tj

dv/ dt =3ti+4tj

v= dr/dt= 3t^2/2+5 i 2t^2 + 2 j

r= (t^3/2+5t +20 )i + (2t^3/3 + 2t +40) j

angle is arctan( (2t^3/3 + 2t +40)/ (t^3/2+5t +20 )i

2. t =sqrt 2s/a=sqrt(29/2)

a. y =ut +0.5at^2=8sqrt(29/2)+29/2=44.96m

speed =sqrt(Vx^2+Vy^2) =sqrt(4*3.8^2+(8+2*3.8)^2)=21.7m/sec

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