Respuesta :
Answer:
a ave = ( v1x - v0x ) i/Δt + v1y j/Δt
Explanation:
The acceleration of a body is the rate of change of the velocity of the body
The correct equation and value for the average acceleration vector and magnitude of the acceleration of the car are as follows;
Part (a) The average acceleration vector is [tex]\overset \rightarrow a_{av} =\dfrac{(v_{1x} - v_{0x})}{\Delta t} }\mathbf{i} + \dfrac{v_{iy} }{\Delta t} \mathbf{j}[/tex]
Part (b) The magnitude of the cars acceleration is [tex]\mathbf{ 2.\overline{518} }[/tex] m/s²
The reason the above values are correct is as follows:
Question: The parts of the question that appears to be missing are;
Part (a); The vector expression for the average acceleration in the time periods
Part(b); Find the magnitude of the cars acceleration
Variables:
v₀ₓ = 31 m/s
v₁ₓ = 31 m/s
[tex]v_{1y}[/tex] = 6.8 m/s
Δt = 2.7 s
Solution:
Part (a) The average acceleration vector, [tex]\mathbf{\overset \rightarrow a_{av}}[/tex], is given as follows;
[tex]Average \ acceleration \ vector, \overset \rightarrow a_{av} = \mathbf{\dfrac{\overset \rightarrow v_2 - \overset \rightarrow v_1}{t_2 - t_1}} =\dfrac{\Delta \overset \rightarrow v}{\Delta t}[/tex]
The given parameters in the question;
v₀ = v₀ₓ i
v₁ = v₁ₓ i + [tex]v_{1y}[/tex]j
∴ v₁ - v₀ = v₁ₓ i + [tex]v_{1y}[/tex]j - v₀ₓ i
The average acceleration vector becomes;
[tex]\overset \rightarrow a_{av} = \mathbf{\dfrac{\overset \rightarrow v_1 - \overset \rightarrow v_0}{\Delta t}} =\dfrac{(v_{1x} - v_{0x})\mathbf{i} + v_{iy} \mathbf{j}}{\Delta t}[/tex]
Therefore;
[tex]\mathbf{The \ average \ acceleration \ vector} \overset \rightarrow a_{av} =\dfrac{(v_{1x} - v_{0x})}{\Delta t} }\mathbf{i} + \dfrac{v_{iy} }{\Delta t} \mathbf{j}[/tex]
Part (b) The magnitude of the car's acceleration is given by plugging values of the variables v₀ₓ, v₁ₓ, [tex]\mathbf{v_{1y}}[/tex], and Δt in the above equation as follows;
[tex]\overset \rightarrow a_{av} =\dfrac{(31 \, m/s - 31 \, m/s)\mathbf{i} + 6.8 \, m/s \mathbf{j}}{2.7 \, s} = 2.\overline{518} \ m/s^2 \ \mathbf{j}[/tex]
The magnitude of the cars acceleration, a = [tex]\sqrt {(2.\overline{518} \ m/s)^2 } = \mathbf{ 2.\overline{518} \ m/s^2}[/tex]
Learn more about acceleration vector here:
https://brainly.com/question/13378384
