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.

Question

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 :)

Answer 1

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.

Answer 2

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.