Question

You're driving down the highway late one night at 20m/s when a deer steps onto the road 60 m in front of you. Your reaction time before stepping on the brakes is 0.50s, and the maximum deceleration of your car is 10m/s2.

Answer 1

If she is driving at 20 m/s when suddenly she sees a deer in the road 60 m in front of her , the car will stop before it hits the deer.

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:

Reaction Time = t' = 0.50 s

Acceleration = a = -10 m/s²

Initial Velocity = u = 20 m/s

Final Velocity = v = 0 m/s

Unknown:

Distance = d = ?

Solution:

When driver sees a deer, she has a reaction time of 0.50 s. It means that it takes 0.50 s before the car start to decelerate. We could calculate the car's distance during this time as shown below :

$Distance ~ During ~ Reaction ~ Time = d' = u \times t'$

$d' = 20 \times 0.5$

$\boxed {d' = 10~m}$

The time needed to slow down the car until it stops could be calculated as shown below :

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

$-10 = \frac{0 - 20}{t}$

$-10 = \frac{-20}{t}$

$t = \frac{-20}{-10}$

$\boxed {t = 2 ~s}$

The distance of the car during deceleration could be calculated as shown below :

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

$d = \frac{20 + 0}{2} \times 2$

$d = 10 \times 2$

$\boxed {d = 20 ~ m}$

At last , the total distance of the car from the moment the driver sees the deer is :

$Total ~ Distance = d + d'$

$Total ~ Distance = 20 ~ m + 10 ~ m$

$\large {\boxed {Total ~ Distance = 30 ~ m} }$

Total Distance is less than 60 m , therefore the car will stop before it hits the deer.

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

Answer 2

Traveling speed = 20 m/s

Within the reaction time of 0.050 s, the distance traveled is
s₁ = (20 m/s)*(0.5 s) = 10 m

The time required to stop with a deceleration of 10 m/s² is
t = (20 m/s)/(10 m/s²) = 2 s.

The distance traveled before coming to a stop is
s₂ = (20 m/s)*(2 s) - 0.5*(10 m/s²)*(2 s)²
    = 20 m

Total distance traveled before coming to a stop is
s₁ + s₂ = 10 + 20 = 30 m

Answer:
This distance of 30 m is less than 60 m. so the vehicle will not hit the deer.