Question

1. -12 points PSE6 4.P.002. My Notes A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by the following expressions. x (20.0 m/s)t y (4.90 m/s)t -(4.76 m/s2)t2 (a) Write a vector expression for the balls position as a function of time, using the unit vectors i and j. (Use the following as necessary: t.) m) j (b) By taking derivatives, obtain an expression for the velocity vector v as a function of time m/s) i m/s) j (c) By taking derivatives, obtain an expression for the acceleration vector a as a function of time m/s2) i m/s2) j (d) Next use unit vector notation to write an expression for the position of the golf ball at t 3.10 s r(3.10 s)- (e) Write an expression for the velocity at this time v(3.10 s) - (f) Write an expression for the acceleration at this time a(3.10 s)- m)j m/s)j m/s2) j If the number is a decimal please include 4 digits past the decimal point if available.

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Answer #1

Given

x and y coordinate as:

x= (20.0 m/s) * t

y = ( 4.90 m/ s) * t - (4.76m/ s^2) *t^2

Part ( a )

position vector ( r ) is given as :

r = ​​​​​​x i + y j

r=(20* t m) i + {( 4.9* t - 4.76 * t^2 )m} j ......... eqn 1)

here m = meter

part (b)

v= deriveitive of position vector (r) w.r.t

v= d r/dt

v= 20 m/s i + {(4.90 -4.76 *2* t )m/s } j

v= 20m/s​​​​​​ i​​​​​​ + {(4.90-9.52*t) m/s} j ............... eqn 2)

part ( C)

a = dv /dt

a = 0 m/s^2 i + {(0 - 9.52)m/s^2} j

a =( -9.52 m/ s^2) j ...........eqn 3)

Part (d)

position vector of golf at t= 3.10m/s

r=( 20* 3.10) i + (4.90 * 3.10- 4.76 * 3.10^2) j

r=( 62 m) i +{( 15.19- 45.7436)m} j

r = (62m) i - {(30. 5536)m} j

part (e)

velocity vector at t= 3.10 s

v = 20 i + ( 4.90 - 9.52* 3.10) j ...... ( from eqn 2 )

v= (20m/s) i - {(24.612 ) m/ s } j

part (f)

a ( at t= 3.10 s)   from eqn 3 we have

a=( - 9.52 m/s^2) j ( it is independent of time)

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