Answer:
The stopping distance would be 200 m.
Explanation:
Hi there!
The work done to stop the vehicle is equal to its change in kinetic energy.
The equation of kinetic energy is the following:
KE = 1/2 · m · v²
Where:
KE = kinetic energy.
m = mass.
v = velocity.
The change in kinetic energy is calculated as follows:
ΔKE = final kinetic energy - initial kinetic energy
In this case, the vehicle is brought to stop, so, the final kinetic energy will be zero.
ΔKE = 0 - 1/2 · m · v²
The work done is calculated as follows:
W = F · d
Where:
W = work done
F = applied force
d = traveled distance (stopping distance in this case)
The force F is calculated as follows:
F = m · a
Where:
m = mass
a = acceleration
Then:
W = ΔKE
F · d = -1/2 · m · v²
m · a · d = -1/2 · m · v²
a · d = -1/2 · v²
d = -1/2 · v² / a
Let´s find the acceleration of the vehicle that is brought to stop in 50 m with an initial velocity of 45 km/h.
Let´s convert 45 and 90 km/h into m/s
45 km/h · 1000 m/ 1 km · 1 h /3600 s = 12.5 m/s
90 km/h · 1000 m/ 1 km · 1 h /3600 s = 25 m/s
The distance and velocity of the vehicle is calculated using the following equations:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = traveled distance at time t.
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
v = velocity at time t.
Let´s place the origin of the frame of reference at the point where the vehicle begins to decelerate so that x0 = 0. When the vehicle stops, its velocity is zero. Let´s use the equation of velocity to find the time it takes the vehicle to stop (and travel a distance of 50 m):
v = v0 + a · t
0 = 12.5 m/s + a · t
-12.5 m/s / t = a
Using the equation of traveled distance, let´s find the time it takes the vehicle to travel 50 m until stop:
x = x0 + v0 · t + 1/2 · a · t²
Replacing a = -12.5 m/s / t
50 m = 12.5 m/s · t + 1/2 · (-12.5 m/s/t) · t²
50 m = 12.5 m/s · t - 6.25 m/s · t
50 m = 6.25 m/s · t
50 m/ 6.25 m/s = t
t = 8.0 s.
Then, the acceleration is the following:
-12.5 m/s / t = a
-12.5 m/s / 8 s = a
a = -1.5625 m/s²
Then, the stopping distance of the vehicle if it travels at an initial speed of 90 km/h would be the following:
d = -1/2 · v² / a
d = -1/2 ·(25 m/s)² / -1.5625 m/s²
d = 200 m
The stopping distance would be 200 m.
