Question

Relative to the ground, a car has a velocity of 16.7 m/s, directed due north. Relative to this car, a truck has a velocity of 25.0 m/s, directed 52.0° north of east. What is the magnitude of the truck's velocity relative to the ground?

Answer 1

Relative to the ground, a car has a velocity of $16.7\mathrm{m / s}$ , directed due north. Relative to this car, a truck has a velocity of $25.0\mathrm{m / s}$ , directed $52.0{}^{\circ}$ north of east. What is the magnitude of the truck's velocity relative to the ground?

Solution:

$V_{t} -$ velocity of the truck relative to the ground;

$V_{t,c} = 25.0\frac{m}{s} -$ velocity of the truck relative to the car;

$V_{c} = 16.7\frac{m}{s} -$ velocity of the car relative to the ground;

$\alpha = 52.0{}^{\circ} -$ the angle between the west

- east direction and the direction of the truck

Formula for the relative velocity of the track:

$$\begin{array}{l} \vec {V} _ {\mathrm {t}, \mathrm {c}} = \vec {V} _ {\mathrm {t}} - \vec {V} _ {\mathrm {c}} \\ \vec {V _ {t}} = \vec {V} _ {t, c} + \vec {V _ {c}} \\ \end{array}$$

The cosine theorem in the triangle ABC $(\beta = 180{}^{\circ} - (90{}^{\circ} - \alpha) = 90{}^{\circ} + \alpha)$ :

$$\begin{array}{l} V _ {t} = \sqrt {V _ {t , c} ^ {2} + V _ {c} ^ {2} - 2 V _ {t , c} \cdot V _ {c} \cdot \cos \beta} = \sqrt {V _ {t , c} ^ {2} + V _ {c} ^ {2} - 2 V _ {t , c} \cdot V _ {c} \cdot \cos (9 0 {}^ {\circ} + \alpha)}; \\ \cos (9 0 {}^ {\circ} + \alpha) = - \sin \alpha \Rightarrow \\ V _ {t} = \sqrt {\left(1 6 . 6 \frac {m}{s}\right) ^ {2} + \left(2 5 \frac {m}{s}\right) ^ {2} + 2 \left(1 6 . 6 \frac {m}{s}\right) \cdot \left(2 5 \frac {m}{s}\right) \cdot \sin 5 2 {}^ {\circ}} = 3 9. 4 3 \frac {m}{s} \\ \end{array}$$

Answer: magnitude of the truck's velocity relative to the ground is $39.43\frac{\mathrm{m}}{\mathrm{s}}$