Question

An engineer in a locomotive sees a car stuck

on the track at a railroad crossing in front of
the train. When the engineer first sees the
car, the locomotive is 350 m from the crossing
and its speed is 23 m/s.
If the engineer's reaction time is 0.54 s,
what should be the magnitude of the mini-
mum deceleration to avoid an accident?
Answer in units of m/s².

(part 1 of 3)
A tennis ball is dropped from 1.54 m above
the ground. It rebounds to a height of 1.03 m.
With what velocity does it hit the ground?
The acceleration of gravity is 9.8 m/s². (Let
down be negative.)
Answer in units of m/s.

(part 2 of 3)
With what velocity does it leave the ground?
Answer in units of m/s.
(part 3 of 3)
If the tennis ball were in contact with the
ground for 0.00767 s, find the acceleration
given to the tennis ball by the ground.

Answer 1

(part 1 of 3)
To find the velocity at which the tennis ball hits the ground, we can use the equation:

v² = u² + 2as

where v is the final velocity (unknown), u is the initial velocity (0 m/s since the ball is dropped), a is the acceleration due to gravity (-9.8 m/s²), and s is the distance (1.54 m).

Rearranging the equation:

v² = 0² + 2(-9.8)(1.54)
v² = -29.9928

Taking the square root of both sides:

v ≈ -5.47 m/s

Therefore, the velocity at which the tennis ball hits the ground is approximately 5.47 m/s downward.

(part 2 of 3)
Since the ball rebounds to a height of 1.03 m, we can use the equation from part 1 to find the velocity with which it leaves the ground.

Using the same equation:

v² = u² + 2as

where v is the final velocity (unknown), u is the initial velocity (unknown), a is the acceleration due to gravity (-9.8 m/s²), and s is the distance (1.03 m).

Rearranging the equation:

v² = u² + 2(-9.8)(1.03)
v² = -20.1872

Taking the square root of both sides:

v ≈ -4.49 m/s

Therefore, the velocity with which the tennis ball leaves the ground is approximately 4.49 m/s upward.

(part 3 of 3)
To find the acceleration given to the tennis ball by the ground, we can use the equation:

a = (v - u) / t

where a is the acceleration (unknown), v is the final velocity (4.49 m/s), u is the initial velocity (0 m/s), and t is the time the ball is in contact with the ground (0.00767 s).

Plugging in the values:

a = (4.49 - 0) / 0.00767
a ≈ 585.21 m/s²

Therefore, the acceleration given to the tennis ball by the ground is approximately 585.21 m/s².