The coordinates of a bird flying in the xy-plane are given by \( x(t) = \alpha t \) and \( y(t) = 3.0 \, \text{m} - \beta t^2 \), where \( \alpha = 2.4 \, \text{m/s} \) and \( \beta = 1.2 \, \text{m/s}^2 \).
a) Let the direction be the angle that the vector makes with the +x-axis measured counterclockwise. Calculate the direction of the bird's velocity at \( t = 2.0 \, \text{s} \). Express your answer in degrees using two significant figures.
b) Calculate the magnitude of the bird's acceleration at \( t = 2.0 \, \text{s} \). Express your answer using two significant figures.
Answer (3)
The **direction **of the bird's **velocity at t=2.0s is ** -63.44°
The magnitude of the bird's acceleration at t=2.0s is -2.4 m/s².
What is acceleration?
Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of **acceleration **is meter/second² (m/s²).\t
The **position **of the bird is:
x(t)=αt
y(t)=3.0m−βt²
The x component of **velocity **is: v₁ =dx/dt =α
The y component of **velocity **is: v₂ =dy/dt =- 2βt
The x component of **acceleration **is: a₁ = dv₁/dt = 0
The y component of **acceleration **is: a₂ = dv₂/dt = -2β
Hence, the **direction **of the bird's **velocity **at t=2.0s is = tan⁻¹(v₂/v₁)
= tan⁻¹(- 2βt/α)
= tan⁻¹(- 2×1.2×2/2.4)
= tan⁻¹(- 2)
= - 63.44°
Hence, the **magnitude **of the bird's acceleration at t=2.0s is = -2β
= - 2 × 1.2 m/s².
= - 2.4 m/s².
Learn more about **acceleration **here:
brainly.com/question/12550364
#SPJ6
magnitude of accel = 2.4m/s2 angle 296.57 degrees , sry I'm new , how do I know if I'm right or not
The direction of the bird's velocity at t = 2.0 s is approximately − 6 3 ∘ and the magnitude of the bird's acceleration at that time is 2.4 m/s 2 .
;
