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 : https://brainly.com/question/3813437
- Kinetic Energy : https://brainly.com/question/692781
- Acceleration : https://brainly.com/question/2283922
- The Speed of Car : https://brainly.com/question/568302
Answer details
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate
