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.
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' = 0.5 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 :
[tex]Distance ~ During ~ Reaction ~ Time = d' = u \times t'[/tex]
[tex]d' = 20 \times 0.5[/tex]
[tex]d' = 10~m[/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]-6 = \frac{0 - 20}{t}[/tex]
[tex]-6 = \frac{-20}{t}[/tex]
[tex]t = \frac{-20}{-6}[/tex]
[tex]t = \frac{10}{3}~s[/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{20 + 0}{2}~\frac{10}{3}[/tex]
[tex]d = 10 \times \frac{10}{3}[/tex]
[tex]d = \frac{100}{3}~m[/tex]
At last , the total distance of the car from the moment the driver sees the obstacle is :
[tex]Total ~ Distance = d + d'[/tex]
[tex]Total ~ Distance = \frac{100}{3}~m + 10~m[/tex]
[tex]Total ~ Distance = \frac{130}{3}~m[/tex]
[tex]\large {\boxed {Total ~ Distance \approx 43~m} }[/tex]
Total Distance is less than 50 m , therefore the car will stop before it hits the obstacle.
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle
