Question

A car is driven 225 km west and then 98 km southwest (45).

Answer 1

Displacement of the car is about 302 km at 13° south of west

Further explanation

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem !

Given :

d₁ = 225 km west

d₂ = 98 km southwest

Unknown :

displacement = d = ?

Solution :

$d^2 = (d_1)^2 + (d_2)^2 + 2d_1d_2\cos 45^o$

$d^2 = (225)^2 + (98)^2 + 2(225)(98)\frac{1}{2}\sqrt{2}$

$d^2 = 60229 + 22050 \sqrt{2}$

$\large {\boxed {d \approx 302 ~km} }$

$\cos \theta = \frac{(d_1)^2+d^2-(d_2)^2}{2d_1d}$

$\cos \theta = \frac{225^2+302^2-98^2}{2(225)(302)}$

$\cos \theta \approx 0.974$

$\large {\boxed {\theta \approx 13^o} }$

Conclusion :

Displacement of the car is about 302 km at 13° south of west

Learn more

• Velocity of Runner : brainly.com/question/3813437
• Kinetic Energy : brainly.com/question/692781
• Acceleration : brainly.com/question/2283922
• The Speed of Car : brainly.com/question/568302

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate

Answer 2

Sum the vector components: Dx = 225* Cos(180)+ 78*Cos(225)= -280.154 km Dy = 225* Sin(180)+ 78*Sin(225) = -55.154 km displacement: Sqrt(Dx^2+Dy^2) = 285.532 km Arctan(Dy/Dx) = 191.137degrees CCW OR: 11.137 degrees South of West