1. A car traveling at a constant speed of 35.2 m/s passes a trooper hidden behin

Question

1. A car traveling at a constant speed of 35.2 m/s passes a trooper hidden behind a billboard. One second later the trooper starts the car with a constant acceleration of 3.99 m/s 2 . How long after the trooper starts the chase does he overtake the speeding car? Answer in units of s. Also to the proper sig figs thanks.

2. 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 100 m from the crossing and its speed is 15 m/s. If the engineer’s reaction time is 0.44 s, what should be the magnitude of the mini-mum deceleration to avoid an accident? Answer in units of m/s 2 . Also to the proper sig figs.

3. A toy rocket, launched from the ground, rises vertically with an acceleration of 12 m/s 2 for 8.2 s until its motor stops. Disregarding any air resistance, what maximum height above the ground will the rocket achieve? The acceleration of gravity is 9.8 m/s 2 . Answer in units of km. Also to the proper sig figs. thanks.

Explanation / Answer

let t represent the time when the car passes the trooper. The troop starts to accelerate 1 second later. t - 1 is the time at which the troop stars to accelerate

the motion of the car is constant
Xf = vt
Xf = 35.2t

the motion of the troop is accelerating
Xf = .5at^2 + Vt + Xi
Xf = .5(4)t^2 + 0t + 0
Xf = 2t^2

when the trooper catches the car, they are in a same position
Trooper = car

2t^2 = 35.2t
1.5t^2 - 24t = 0
t (1.5t - 24) = 0
t = 17.6 seconds

recall that the trooper starts to accelerate 1 second later. So 17.6 - 1 = 16.6 s is the take it takes the trooper to catch the car.

16.6 seconds

2)

Let’s determine the distance the train moves during the engineer’s reaction time.
d = 15 * 0.44 = 6.6 m
100-6.6 = 93.4 m
This is the distance between the train and car, when the train begins to decelerate. Use the following equation to determine the minimum deceleration.

vf^2 = vi^2 + 2 * a * d
0 = 15^2 + 2 * a * 93.4
-15^2 = a * 186.8
a = - 0.83 m/s2
3)

vi = 0
vf = vi + a*t = 35 m/s^2 * 8.6 s = 301 m/s

2) Find the change in height of the rocket under acceleration

vf^2 = vi^2 + 2*a*h
vf^2 = 0 + 2*a*h
h = vf^2 / (2*a) = (301 m/s)^2 / (2*35 m/s^2) = 1294.3 m

3)

a)

Find the velocity at the end of the acceleration.

vi = 0
vf = vi + a*t = 12 m/s^2 * 8.2 s = 98.4 m/s

b) Find the change in height of the rocket under acceleration

vf^2 = vi^2 + 2*a*h
vf^2 = 0 + 2*a*h
h = vf^2 / (2*a) = (98.4 m/s)^2 / (2*12 m/s^2) = 403.44 m

c) Find how long the rocket takes to quit rising under gravity

vi = 98.4 m/s
vf = 0 m/s

vf = vi - g*t
0 = vi - g*t

t = vi/g = 98.4 m/s / 9.81 m/s^2 = 10.03 s

d) Use the equation of motion to find the total height

hf = hi + vi*t - 1/2*g*t^2

hi = 403.44
vi = 98.4 m/s
t = 10.03 s

hf = 403.44 m + 98.4 m/s * 10.03 s - 1/2 * 9.81 m/s^2 * (10.03 s)^2

hf = 1390.392-493.44

hf = 896.94 m = 0.896 km

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