A particle starts from the origin at t = 0 s with a velocity of 10.0 j m/s and moves in the x-y plane with a constant acceleration of (8.0 i + 2.0j ) m s-2.
(a) At what time is the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?
(b) What is the speed of the particle at the time ?
For motion with constant acceleration:
\( x(t) = v_{0x} t + \dfrac{1}{2} a_{x} t^{2} \) \( y(t) = v_{0y} t + \dfrac{1}{2} a_{y} t^{2} \)
Here, \( v_{0x} = 0 \), \( a_{x} = 8.0 \,\text{m s}^{-2} \), so
\( x(t) = 0 + \dfrac{1}{2} (8.0) t^{2} = 4 t^{2} \)
Set \( x = 16 \,\text{m} \):
\( 4 t^{2} = 16 \Rightarrow t^{2} = 4 \Rightarrow t = 2.0 \,\text{s} \) (take positive root)
With \( v_{0y} = 10.0 \,\text{m s}^{-1} \), \( a_{y} = 2.0 \,\text{m s}^{-2} \):
\( y(t) = 10.0\, t + \dfrac{1}{2} (2.0) t^{2} = 10 t + t^{2} \)
\( y(2) = 10 \times 2 + (2)^{2} = 20 + 4 = 24 \,\text{m} \)
(a) Time when \( x = 16 \,\text{m} \): \( t = 2.0 \,\text{s} \); corresponding \( y \)-coordinate: \( y = 24 \,\text{m} \).
Velocity with constant acceleration:
\( v_{x}(t) = v_{0x} + a_{x} t = 0 + 8.0 \, t \) \( v_{y}(t) = v_{0y} + a_{y} t = 10.0 + 2.0 \, t \)
At \( t = 2.0 \,\text{s} \):
\( v_{x}(2) = 8.0 \times 2 = 16.0 \,\text{m s}^{-1} \) \( v_{y}(2) = 10.0 + 2.0 \times 2 = 14.0 \,\text{m s}^{-1} \)
Speed is the magnitude of the velocity vector:
\( v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(16.0)^{2} + (14.0)^{2}} \) \( = \sqrt{256 + 196} = \sqrt{452} \approx 21.3 \,\text{m s}^{-1} \)
(b) Speed of the particle at \( t = 2.0 \,\text{s} \): \( v \approx 21.3 \,\text{m s}^{-1} \).