Answer to Question #12283 in Mechanics | Relativity for aneequa

Question

a truck is moving forward at a constant speed of 21m/s. the driver sees a stationery car directly ahead at a distance of 110m. after a reaction time of delta t he applies the brakes, which gives the truck an acceleration of -3m/s*s a)what is the maximum allowable delta t to avoid collision and what distance will the truck have moved before the brakes take hold ? b0 assuming a reaction of time of 1.4 sec show far behind the car will the truck stop, and in how many seconds from the time the driver first saw the car?

Answer 1

a) what is the maximum allowable delta t to avoid collision and what
distance
will the truck have moved before the brakes take hold?
V0 + a * ts = 0
ts
= V0 / (-a)
ts = 21 / 3 = 7 s (the time needed to stop the truck)
Braking
distance: S(ts) = V0* ts + a * ts^2 / 2 = 21 * 7 - 3 * 49 / 2
= 147 - 73.5 =
73.5 m
deltaS = D - S(ts) = 110 - 73.5 = 36.5 m
deltaS = V0 *
deltat
deltat = deltaS / V0
deltat = 36.5 / 21 = 1.73 s
b) assuming a
reaction of time of 1.4 sec show far behind the car will
the truck stop, and
in how many seconds from the time the driver first
saw the car?
Let deltat
= 1.4 s.
l = D - S(ts) - V0* deltat
l = 110 - 73.5 - 21 * 1.4 = 7.1 m
t
= deltat + ts
t = 1.4 + 7 = 8.4 s
Answer:
a) 1.73 s
b) 7.1 m; 8.4
s.