Answer:
(a) Distance between deer and car = 5 m
(b) Vmax = 21.92 m/s
Explanation:
a.
First we calculate distance covered during response time:
s₁ = vt --------- equation 1
where,
s₁ = distance covered during response time = ?
v = speed of car = 20 m/s
t = response time = 0.5 s
Therefore,
s₁ = (20 m/s)(0.5 s)
s₁ = 10 m
Now, we calculate the distance covered by the car during deceleration. Using 3rd equation of motion:
2as₂ = Vf² - Vi²
s₂ = (Vf² - Vi²)/2a ------ eqation 2
where,
a = deceleration = - 10 m/s²
s₂ = Distance covered during deceleration = ?
Vf = Final Velocity = 0 m/s (since car finally stops)
Vi = Initial Velocity = 20 m/s
Therefore,
s₂ = [(0 m/s)² - (20 m/s)²]/2(-10 m/s²)
s₂ = (400 m²/s²)/(20 m/s²)
s₂ = 20 m
thus, the total distance covered by the car before coming to rest is given as:
s = s₁ + s₂
s = 10 m + 20 m
s = 30 m
Now, the distance between deer and car, when it comes to rest, can be calculated as:
Distance between deer and car = 35 m - s = 35 m - 30 m
<u>Distance between deer and car = 5 m</u>
b.
Since, the distance covered by the car in total must be equal to 35 m at maximum. Therefore,
s₁ + s₂ = 35 m
using equation 1 and equation 2 from previous part:
Vi t + (Vf² - Vi²)/2a = 35 m
Vi(0.5 s) + [(0 m/s)² - Vi²]/2(-10 m/s²) = 35 m
0.5 Vi + 0.05 Vi² = 35
0.05 Vi² + 0.5 Vi - 35 = 0
solving this quadratic equation, we get:
Vi = - 31.92 m/s (OR) Vi = 21.92 m/s
For maximum velocity:
<u>Vmax = 21.92 m/s</u>
