Answer:
Explanation:
From the given information:
The car's initial velocity = 4 m/s in the direction of east 45° due north
We can therefore express the vector of this component form as:
v₁ = (4 m/s) (cos(45º)i + sin(45º)j)
v₁ = (2.83 m/s)i + (2.83 m/s)j
Similarly, the car's final velocity = 4 m/s in the direction of the east side 10º north
∴
v₂ = (4 m/s) (cos(10º) i + sin(10º) j)
v₂ = (3.94 m/s) i + (0.695 m/s) j
From the first equation of motion
v = u + at
Making acceleration "a" the subject of the formula, we have:
a = (v - u )/t
a = (v₂ - v₁)/t
a = (0.370 m/s²) + (-0.711 m/s²)
The magnitude of the avg. acceleration is:
[tex]|| a||= \sqrt{(0.370 m/s^2)^2 + (-0.711 m/s^2)^2)[/tex]
[tex]|| a||= 0.8015 \ m/s^2[/tex]
And;
The direction can be determined by taking the tangent of the acceleration:
i.e.
[tex]tan(\theta) = \dfrac{-0.711 m/s^2}{ 0.370 m/s^2}[/tex]
[tex]tan(\theta) = -1.9216[/tex]
[tex]\theta = tan^{-1} ( -1.9216 )[/tex]
[tex]\mathbf{\theta = -62.51 ^0}[/tex]
Thus, the direction of the angle is approximately S 62.51º E
