Answer:
The second driver travels a farther distance than the first driver when the respective brakes are applied.
Explanation:
Let the first driver be driver A and the second driver be driver B
Let the time brake was applied in the two cases be t = 0
The sketch of the v-t diagram of the two drivers is presented on the attached image to this solution.
The graph is negative sloping because acceleration is negative and the cars are decelerating to a stop.
The distance covered by the two drivers can be obtained from the area under their respective v-t graphs, or from using the equations of motion.
Area under the two graphs is the area of a triangle, (bh/2)
For driver A
Distance covered = (1/2)×25×10 = 125 m
For Driver B
Distance covered = (1/2)×15×20 = 150 m
We can then use the equations of motion to obtain the distances covered by the drivers during the braking period.
For driver A
u = initial velocity = 25 m/s
v = final velocity = 0 m/s (since the car comes to rest)
t = 10 s
a = acceleration = ?
x = distance covered = ?
v = u + at
0 = 25 + 10a
a = (-25/10) = -2.5 m/s²
x = ut + (1/2)at²
x = 25×10 + (1/2)×-2.5×10²
x = 250 - 125 = 125 m
For driver B
u = initial velocity = 15 m/s
v = final velocity = 0 m/s (since the car comes to rest)
t = 20 s
a = acceleration = ?
x = distance covered = ?
v = u + at
0 = 15 + 20a
a = (-15/20) = -0.75 m/s²
x = ut + (1/2)at²
x = 15×20 + (1/2)×-0.75×20²
x = 300 - 150 = 150 m
So, it is evident that the second driver travels a farther distance than the first driver when the respective brakes are applied.
Hope this Helps!!!
