Question

A driver has a reaction time of 0.50 s , and the maximum deceleration of her car is 6.0 m/s2 . she is driving at 20 m/s when suddenly she sees an obstacle in the road 50 m in front of her.

Answer 1

If she is driving at 20 m/s when suddenly she sees an obstacle in the road 50 m in front of her , the car will stop before it hits the obstacle.

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' = 0.5 s

Acceleration = a = -6 m/s²

Initial Velocity = u = 20 m/s

Final Velocity = v = 0 m/s

Unknown:

Distance = d = ?

Solution:

When driver sees an obstacle , she has a reaction time of 0.5 s. It means that it takes 0.5 s before the car start to decelerate. We could calculate the car's distance during this time as shown below :

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

$d' = 20 \times 0.5$

$d' = 10~m$

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

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

$-6 = \frac{0 - 20}{t}$

$-6 = \frac{-20}{t}$

$t = \frac{-20}{-6}$

$t = \frac{10}{3}~s$

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

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

$d = \frac{20 + 0}{2}~\frac{10}{3}$

$d = 10 \times \frac{10}{3}$

$d = \frac{100}{3}~m$

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

$Total ~ Distance = d + d'$

$Total ~ Distance = \frac{100}{3}~m + 10~m$

$Total ~ Distance = \frac{130}{3}~m$

$\large {\boxed {Total ~ Distance \approx 43~m} }$

Total Distance is less than 50 m , therefore the car will stop before it hits the obstacle.

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

I will assume that you want to know if the driver can stop before hitting the obstacle or not.

First step is to use the reaction time in order to know how far can she go before her motion starts to slow:
distance = velocity x time = 20 x 0.5 = 10 m
Thus, the driver has 50 - 10 = 40 m to stop before colliding

Second step is to calculate the distance that the driver requires to stop using the rules of velocity and distance:
1- velocity = acceleration x time 
     20 = 6t ..............> thus, time = 3.334 seconds
2- distance = 0.5 x a x t^2 = 0.5 x 6 x (3.334)^2 = 33.3466 m

From the previous calculations, we can see that the driver has 40 m to stop and she needs only 33.3466 m to stop based on the given parameters. This means that she can stop before collision.