The driver of a car moving at 25m/s applies breakes, decelerates uniformly and stops at 10s. Another driver moving at 15m/s applies breakes, decelerates uniformly and stops at 20s. Plot a v-t graph for the 2 cars. Which of the 2 cars travelled farther after breaks were applied

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!!!

answer: i have one answer

acids are substances that ionize (break off) in an aqueous solution to produce hydrogen (h+) ions

explanation:

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