Answer :
Have you considered hitting it one piece at a time?
Reaction Distance: Speed * 0.50 s
Initial Problem: 16 m/s * 0.50 s = 8 m
Braking Acceleration: a(t) = -10 m/s^2 --
This is constant.
Braking Velocity: s(t) = -10 m/s^2 (t) + "Initial Velocity"
Initial Problem: s(t) = -10 m/s^2 (t) + 16 m/s
Braking Location: x(t) = -5 m/s^2 (t^2) + 16 m/s (t) - "Initial Location"
Initial Problem: x(t) = -5 m/s^2 (t^2) + 16 m/s (t) - (44 m - 8 m)
Having said all that, the first question makes no sense. Does it mean the distance when the vehicle stops?
s(t) = -10 m/s^2 (t) + 16 m/s = 0 ==> t = 16/10 s = 8/5 s
That's how long it takes to stop.
Where are we when that happens? x(8/5) = -5 m/s^2 ((8/5 s)^2) + 16 m/s (8/5 s) - (44 m - 8 m) = -23.2 m
This seems to be a little different from your response. Now, the challenge is to do this all over again, but missing the initial velocity.
Initial Velocity: V -- Let's just call it this so we can talk about it.
Initial Distance is the same: 44 m -- No change.
Reaction Distance: V/2
Braking Acceleration: a(t) = -10 m/s^2 --
No Change Braking Velocity: s(t) = -10 m/s^2 (t) + V -- Missing V!
Braking Location: x(t) = -5 m/s^2 (t^2) + V (t) - (44 m - V/2 m)
When do we stop? s(t) = -10 m/s^2 (t) + V = 0 ==> t = V/10 s
Where do we stop? x(t) = -5 m/s^2 ((V/10 s)^2) + V (V/10 s) - (44 m - V/2 m) = 0 -- We might just brush the poor, frightened deer.
One piece at a time! Slowly. Methodically.
Reaction Distance: Speed * 0.50 s
Initial Problem: 16 m/s * 0.50 s = 8 m
Braking Acceleration: a(t) = -10 m/s^2 --
This is constant.
Braking Velocity: s(t) = -10 m/s^2 (t) + "Initial Velocity"
Initial Problem: s(t) = -10 m/s^2 (t) + 16 m/s
Braking Location: x(t) = -5 m/s^2 (t^2) + 16 m/s (t) - "Initial Location"
Initial Problem: x(t) = -5 m/s^2 (t^2) + 16 m/s (t) - (44 m - 8 m)
Having said all that, the first question makes no sense. Does it mean the distance when the vehicle stops?
s(t) = -10 m/s^2 (t) + 16 m/s = 0 ==> t = 16/10 s = 8/5 s
That's how long it takes to stop.
Where are we when that happens? x(8/5) = -5 m/s^2 ((8/5 s)^2) + 16 m/s (8/5 s) - (44 m - 8 m) = -23.2 m
This seems to be a little different from your response. Now, the challenge is to do this all over again, but missing the initial velocity.
Initial Velocity: V -- Let's just call it this so we can talk about it.
Initial Distance is the same: 44 m -- No change.
Reaction Distance: V/2
Braking Acceleration: a(t) = -10 m/s^2 --
No Change Braking Velocity: s(t) = -10 m/s^2 (t) + V -- Missing V!
Braking Location: x(t) = -5 m/s^2 (t^2) + V (t) - (44 m - V/2 m)
When do we stop? s(t) = -10 m/s^2 (t) + V = 0 ==> t = V/10 s
Where do we stop? x(t) = -5 m/s^2 ((V/10 s)^2) + V (V/10 s) - (44 m - V/2 m) = 0 -- We might just brush the poor, frightened deer.
One piece at a time! Slowly. Methodically.
Answer:
Part a)
d = 23.2 m
Part b)
v = 25.1 m/s
Explanation:
Part a)
As we know that the speed of the car is given as
[tex]v = 16 m/s[/tex]
reaction time of the driver = 0.50 s
now the distance moved by the car during reaction time
[tex]d = velocity \times time[/tex]
[tex]d = 16(0.50) = 8m[/tex]
After this car is decelerated due to brakes with deceleration
[tex]a = -10 m/s^2[/tex]
so the distance moved by the car till it will stop
[tex]v_f^2 - v_i^2 = 2ad[/tex]
[tex]0 - 16^2 = 2(-10)d[/tex]
[tex]d = 12.8 m[/tex]
so total distance moved by the car
[tex]d = 12.8 + 8 = 20.8 m[/tex]
Now the distance between the car and deer is given as
[tex]x = 44 - 20.8[/tex]
[tex]x = 23.2 m[/tex]
Part b)
Let the driver is moving with speed "v"
now the distance moved in reaction time
[tex]d_1 = v(0.50)[/tex]
now after applying brakes the distance moved by it
[tex]v_f^2 - v_i^2 = 2 ad [/tex]
so we have
[tex]0 - v^2 = 2(-10)d_2[/tex]
so total distance covered is given as
[tex]d = d_1 + d_2[/tex]
[tex]d = \frac{v^2}{20} + 0.50v[/tex]
for maximum speed v the distance moved is
[tex]d = 44 m[/tex]
so we have
[tex]44 = \frac{v^2}{20} + 0.50v[/tex]
by solving above equation we have
[tex]v = 25.1 m/s[/tex]