Question (a)
The cutting of the X axis occurs when the Y coordinate is equal to zero.
The vertical position is given by:Y = Y o + V o y ∗ t + 1 2 a y t 2 Y=Y_{o}+V_{oy}*t+\frac{1}{2}a_{y}t^{2} Y = Y o + V oy ∗ t + 2 1 a y t 2
Where:
Initial position. Y o = 10 m Y_{o}=10m Y o = 10 m Initial velocity V o y = + 5 m s V_{oy}=+5\frac{m}{s} V oy = + 5 s m Vertical acceleration. aa y = − 4 m s 2 a_{y}=-4\frac{m}{s^{2}} a y = − 4 s 2 m Evaluating at Y = 0.
0 = 10 + 5 t − 1 2 4 t 2 − 2 t 2 + 5 t + 10 = 0 0=10+5t-\frac{1}{2}4t^{2}\\ -2t^{2}+5t+10=0 0 = 10 + 5 t − 2 1 4 t 2 − 2 t 2 + 5 t + 10 = 0
A second grade equation.− 2 t 2 + 5 t + 10 = 0 − > a x 2 + b x + c = 0 -2t^{2}+5t+10=0->ax^{2}+bx+c=0 − 2 t 2 + 5 t + 10 = 0 − > a x 2 + b x + c = 0
Coefficients a = − 2 ; b = 5 ; c = 10 a=-2;b=5;c=10 a = − 2 ; b = 5 ; c = 10
Solving the second degree equation.
t = − b ± b 2 − 4 ∗ a ∗ c 2 ∗ a t = − ( 5 ) ± ( 5 ) 2 − 4 ∗ − 2 ∗ 10 2 ∗ − 2 t = − 5 ± 10.2 − 4 t=\frac{-b\pm \sqrt{b^{2}-4*a*c}}{2*a} \\t=\frac{-(5)\pm \sqrt{(5)^{2}-4*-2*10}}{2*-2}\\ t=\frac{-5 \pm 10.2}{-4} t = 2 ∗ a − b ± b 2 − 4 ∗ a ∗ c t = 2 ∗− 2 − ( 5 ) ± ( 5 ) 2 − 4 ∗− 2 ∗ 10 t = − 4 − 5 ± 10.2
Solution 1
t = − 5 + 10.2 − 4 t = − 1.25 s t=\frac{-5 + 10.2}{-4} \\t=-1.25s t = − 4 − 5 + 10.2 t = − 1.25 s
Solution 2
t = − 5 − 10 − 4 t = 3.75 s t=\frac{-5 - 10}{-4} \\t=3.75s t = − 4 − 5 − 10 t = 3.75 s
the time when the particle will cross the x axis
t = 3.75 s \boxed{t=3.75s} t = 3.75 s
Question (b)
The position of the particle is given by:
r ( t ) ⃗ = r ⃗ o + V o ⃗ ∗ t + 1 2 a ⃗ t 2 \vec{r(t)}=\vec{r}_{o}+\vec{V_{o}}*t+\frac{1}{2}\vec{a}t^{2} r ( t ) = r o + V o ∗ t + 2 1 a t 2
Where:
initial velocity V 0 ⃗ = ( 4 i + 5 j ) m s \vec{V_{0}}=(4i+5j)\frac{m}{s} V 0 = ( 4 i + 5 j ) s m The acceleration is a ⃗ = ( 2 i − 4 j ) m s 2 \vec{a}=(2i-4j)\frac{m}{s^{2}} a = ( 2 i − 4 j ) s 2 m Initial position r 0 ⃗ = ( 4 i + 10 j ) m \vec{r_{0}}=(4i+10j)m r 0 = ( 4 i + 10 j ) m time t = 3.75 s t=3.75s t = 3.75 s
Numerically evaluating
r ( t ) ⃗ = ( 4 i + 10 j ) m + ( 4 i + 5 j ) m s ∗ 3.75 s + 1 2 ∗ ( 2 i − 4 j ) m s 2 ∗ ( 3.75 s ) 2 \vec{r(t)}=(4i+10j)m+(4i+5j)\frac{m}{s}*3.75s+\frac{1}{2}*(2i-4j)\frac{m}{s^{2}}*(3.75s)^{2} r ( t ) = ( 4 i + 10 j ) m + ( 4 i + 5 j ) s m ∗ 3.75 s + 2 1 ∗ ( 2 i − 4 j ) s 2 m ∗ ( 3.75 s ) 2
r o ⃗ = ( 4 i + 10 j ) m + ( 15 i + 18 j ) m + ( 14 i − 28 j ) m \vec{r_{o}}=(4i+10j)m+(15i+18j)m+(14i-28j)m r o = ( 4 i + 10 j ) m + ( 15 i + 18 j ) m + ( 14 i − 28 j ) m
r o ⃗ = ( 33 i + 0 j ) m \vec{r_{o}}=(33i+0j)m r o = ( 33 i + 0 j ) m
The position at that moment is:
r o ⃗ = ( 33 i + 0 j ) m or r o ⃗ = ( 33 m , 0 m ) \boxed{\vec{r_{o}}=(33i+0j)m \text{ or } \vec{r_{o}}=(33m,0m)} r o = ( 33 i + 0 j ) m or r o = ( 33 m , 0 m )
Question (c)
The velocity of the particle is given by:
V ( t ) ⃗ = V o ⃗ + a ⃗ ∗ t \vec{V(t)}=\vec{V_{o}}+\vec{a}*t V ( t ) = V o + a ∗ t
Where:
The initial velocityV 0 ⃗ = ( 4 i + 5 j ) m s \vec{V_{0}}=(4i+5j)\frac{m}{s} V 0 = ( 4 i + 5 j ) s m Acceleration a ⃗ = ( 2 i − 4 j ) m s 2 \vec{a}=(2i-4j)\frac{m}{s^{2}} a = ( 2 i − 4 j ) s 2 m Time t=3.75s
Numerically evaluating
V ( t ) ⃗ = ( 4 i + 5 j ) m s + ( 2 i − 4 j ) m s 2 ∗ 3.75 s V ( t ) ⃗ = ( 4 i + 5 j ) m s + ( 5.5 i − 15 j ) m s V ( t ) ⃗ = ( 11.5 i − 10 j ) m s \vec{V(t)}=(4i+5j)\frac{m}{s} +(2i-4j)\frac{m}{s^{2}}*3.75s \\\vec{V(t)}=(4i+5j)\frac{m}{s}+(5.5i-15j)\frac{m}{s}\\\vec{V(t)}=(11.5i-10j)\frac{m}{s} V ( t ) = ( 4 i + 5 j ) s m + ( 2 i − 4 j ) s 2 m ∗ 3.75 s V ( t ) = ( 4 i + 5 j ) s m + ( 5.5 i − 15 j ) s m V ( t ) = ( 11.5 i − 10 j ) s m
The velocity at that moment is
v e c V ( t ) = ( 11.5 i − 10 j ) m s \boxed{vec{V(t)}=(11.5i-10j)\frac{m}{s} } v ec V ( t ) = ( 11.5 i − 10 j ) s m
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