The total distance covered is about 327 ft
Further explanation
Acceleration is rate of change of velocity.
[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]
[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]
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:
Reaction Time = t' = 0.7 s
Acceleration = a = -12 ft/s²
Initial Velocity = u = 55 mi/h = 80.63 ft/s
Final Velocity = v = 0 m/s
Unknown:
Distance = d = ?
Solution:
When driver sees a deer, she has a reaction time of 0.7 s. It means that it takes 0.7 s before the car start to decelerate. We could calculate the car's distance during this time as shown below :
[tex]Distance ~ During ~ Reaction ~ Time = d' = u \times t'[/tex]
[tex]d' = 80.63 \times 0.7[/tex]
[tex]d' = \boxed{56.441 ~ feet}[/tex]
The time needed to slow down the car until it stops could be calculated as shown below :
[tex]a = \frac{v - u}{t}[/tex]
[tex]-12 = \frac{0 - 80.63}{t}[/tex]
[tex]-12 = \frac{-80.63}{t}[/tex]
[tex]t = \frac{-80.63}{-12}[/tex]
[tex]t = \boxed{\frac{80.63}{12} ~ \text{seconds}}[/tex]
The distance of the car during deceleration could be calculated as shown below :
[tex]d = \frac{u + v}{2}~t[/tex]
[tex]d = \frac{80.63 + 0}{2} \times \frac{80.63}{12}[/tex]
[tex]d = 40.315 \times \frac{80.63}{12}[/tex]
[tex]d = \boxed {271 ~ feet}[/tex]
At last , the total distance of the car from the moment the driver sees the deer is :
[tex]Total ~ Distance = d + d'[/tex]
[tex]Total ~ Distance = 271 ~ ft + 56.441 ~ ft[/tex]
[tex]\large {\boxed {Total ~ Distance \approx 327 ~ ft} }[/tex]
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
