The traveled distance of the car is four times as much compared to the truck
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:
Initial velocity of Car = Initial velocity of Truck = u = 0 m/s
Acceleration of Car = Acceleration of Truck = a
Time Taken of Truck = t
Time Taken of Car = 2t
Unknown:
The Travel Distance of the Car : The Travel Distance of the Truck = ?
Solution:
We will compare the travel distance of the car and truck
\(d_{car} : d_{truck} = (ut_{car} + \frac{1}{2}at_{car}^2) : (ut_{truck} + \frac{1}{2}at_{truck}^2)\)
\(d_{car} : d_{truck} = (0(t_{car}) + \frac{1}{2}at_{car}^2) : (0(t_{truck}) + \frac{1}{2}at_{truck}^2)\)
\(d_{car} : d_{truck} = \frac{1}{2}at_{car}^2 : \frac{1}{2}at_{truck}^2\)
\(d_{car} : d_{truck} = \frac{1}{2}at_{car}^2 : \frac{1}{2}at_{truck}^2\)
\(d_{car} : d_{truck} = t_{car}^2 : t_{truck}^2\)
\(d_{car} : d_{truck} = (2t)^2 : (t)^2\)
\(d_{car} : d_{truck} = 4t^2 : t^2\)
\(d_{car} : d_{truck} = 4 : 1\)
\(d_{car} = 4 d_{truck}\)
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
