Question

A mail carrier parks her postal truck and delivers packages. To do so, she walks east at a speed of 0.80 m/s for 4.0 min, then north at a speed of 0.50 m/s for 5.5 min, and finally west at a speed of 1.1 m/s for 2.8 min. Define east as +x and north as +y. (a) Write unit-vector velocities for each of the legs of her journey. (b) Find unit-vector displacements for each of the legs of her journey. (c) Find her net displacement from the postal truck after her journey is complete.

Answer 1

Explanation:

Given that,

She walks in east,

Speed = 0.80 m/s

Time = 4.0 min

In north,

Speed = 0.50 m/s

Time = 5.5 min

In west,

Speed = 1.1 m/s

Time = 2.8 min

(a). We need to calculate the unit-vector velocities for each of the legs of her journey.

The velocity of her in east

$\vec{v_{1}}=0.80\ \hat{x}\ m/s$

$\vec{v_{2}}=0.50\ \hat{y}\ m/s$

$\vec{v_{3}}=1.1\ \hat{-x}\ m/s$

(b). We need to calculate the unit-vector displacements for each of the legs of her journey

Using formula of displacement

$\vec{d_{1}}=v_{1}\times t_{1}$

In east ,

$\vec{d_{1}}=0.80\times4.0\times60$

$\vec{d_{1}}=192\ \hat{x}\ m$

In north,

$\vec{d_{2}}=0.50\times5.5\times60$

$\vec{d_{2}}=165\ \hat{y}\ m$

In west,

$\vec{d_{3}}=1.1\times2.8\times60$

$\vec{d_{3}}=184.8\ \hat{-x}\ m$

(c). We need to calculate the  net displacement from the postal truck after her journey is complete

$\vec{d}=\vec{d_{1}}+\vec{d_{2}}+\vec{d_{3}}$

Put the value in the formula

$\vec{d}=192\hat{x}+165\hat{y}+184.8\hat{-x}$

$\vec{d}=7.2\hat{x}+165\hat{y}$

We need to calculate the magnitude of the displacement

$d=\sqrt{(7.2)^2+(165)^2}$

$d=165.16\ m$

The magnitude of the displacement is 165.16 m.

Hence, This is the required solution.