1.A motorist drive along a straight road at a constant speed of 15.0 m/s.Just as she passes a parked motorcycle police officer,the officer starts to accelerate at 2.0 m/s^2 to overtake her. Assuming the officer maintains this acceleration, a) determine the time it takes the police officer the motorist.

b)the speed,
c)the total displacement of the officer as ge overtakes the motorist.

2.A stone is thrown vertically upward with a speed of 15.0 m/s from the edge of cliff 30.0 m high. At the same instant,another stone is thrown up with a speed of 40.0 m/s frim the foot of the cliff. Will the two stones meet before they hit the ground?give exact details.

thanks :)

1a. d1 = d2

V*t = 0.5a*t^2
15*t = 1*t^2
Divide by t:
t = 15 s.

1b. r = a*t = 2 * 15 = 30 m/s. = Speed
of officer.

1c. D = 15*t = 15 * 15 = 225 m.

2.
1st Stone:
Tr = -Vo/g = -15/-9.8 = 1.53 s. = Rise
time.

h = ho + (Vo*t - 0.5g*t^2)
h = 30 + (15*1.53 - 4.9*1.53^2) =
30 + 22.96 - 11.47 = 41.5 m Above gnd.

0.5g*t^2 = 41.5 m.
4.9t^2 = 41.5
t^2 = 8.47
Tf = 2.91 s. = Fall time.

Tr*Tf = 1.53 + 2.91 = 4.44 s = Time in
air.

2nd Stone:
Tr = -Vo/g = -40/-9.8 = 4.08 s.

h = (V^2-Vo^2)/2g = (0-(40^2)/-19.6 =
81.63 m. Above gnd.

0.5g*t^2 = 81.63
4.9t^2 = 81.63
t^2 = 16.66
Tf = 4.08 s. = Fall time.

Tr+Tf = 4.08 + 4.08 = 8.16 s. In air.

h = 40*1.53 - 4.9*1.53^2 = 61.2 - 11.47 = 49.7 m.

The 1st stone reaches its' maximum ht. of 41.5 m in 1.53 s. But at 1.53 s,
the 2nd stone is at 49.7 m and has passed the 1st stone.

2. Continued.

When the 2nd stone catchup, they will be the same distance above gnd.:

h1 = h2
30 + 15t - 4.9*t^2 = 40*t - 4.9t^2
The 4.9t^2 terms cancel:
30 + 15t - 40t = 0
-25t = -30
t = 1.2 s. To catchup.

h = 40*1.2 - 4.9*1.2^2 = 48 - 7.06 = 40.9 m. Above gnd. = Height at which the
2nd stone passes the 1st stone. Both stones are still rising.

a) To determine the time it takes for the police officer to overtake the motorist, we need to use the equation of motion:

\[d = ut + \frac{1}{2}at^2\]

Where:
- d is the displacement
- u is the initial velocity
- a is the acceleration
- t is the time

In this case, the initial velocity of the police officer is 0 m/s, and the displacement is the same for both the motorist and the police officer since they are starting at the same point.

So, we have:
\[0 = 15t + \frac{1}{2}(2)t^2\]

Rearranging the equation and simplifying:
\[t^2 + 7.5t = 0\]
\[t(t + 7.5) = 0\]

This gives us two possible solutions: t = 0 or t = -7.5. Since time cannot be negative, we discard the negative solution.

Therefore, the time it takes for the police officer to overtake the motorist is t = 0 seconds.

b) To determine the speed at which the police officer overtakes the motorist, we can use the equation:

\[v = u + at\]

where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time

Plugging in the values:
\[v = 0 + 2(0)\]

We get:
\[v = 0\]

The speed at which the police officer overtakes the motorist is 0 m/s.

c) The total displacement of the officer as he overtakes the motorist is equal to the initial displacement of the motorist (since they started at the same point), which is 0 meters.

Therefore, the total displacement of the officer as he overtakes the motorist is 0 meters.

Regarding the second question:

No, the two stones will not meet before they hit the ground. One stone is thrown vertically upward from the edge of the cliff, while the other stone is thrown upwards from the foot of the cliff. Since the stone thrown from the foot of the cliff has a higher initial velocity, it will reach a greater maximum height compared to the stone thrown from the edge of the cliff. Eventually, both stones will start falling down due to gravity, but their paths will not intersect.

Hope this helps!