Why is it that I feel alone
2 answers:
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Missing parts in the text of the exercise:
- The distance from the traffic light is given by
, where 
and 
Solution:
part a) <span>calculate the average velocity of the car for the time interval t=0 to t=8.0 s
- The average velocity is given by the ratio between the distance covered in the time interval:
</span>
x(0 s), the distance covered after t=0 s, is zero, while the distance after t=8.0 s is
Therefore, the average velocity is
part b) <span>calculate the instantaneous velocity of the car at t=0, t=4.0 s and t=8.0 s
- The instantaneous velocity can be found by performing the derivation of x(t):
</span>
<span>So now we just have to substitute t=0, t=4 s and t=8 s:
- t=0: v(0)=0
- t=4 s: </span>
- t=8 s:
part c) <span>how long after starting from rest is the car again at rest?
- To solve this part we must find the value of t for which v(t)=0, so:
</span>
<span>The first solution is t=0 s, which corresponds to the beginning of the motion, so we are not interested in this value. The second solution is
</span>
<span>and this is the time at which the car is at rest again.</span>
- The distance from the traffic light is given by
Solution:
part a) <span>calculate the average velocity of the car for the time interval t=0 to t=8.0 s
- The average velocity is given by the ratio between the distance covered in the time interval:
</span>
x(0 s), the distance covered after t=0 s, is zero, while the distance after t=8.0 s is
Therefore, the average velocity is
part b) <span>calculate the instantaneous velocity of the car at t=0, t=4.0 s and t=8.0 s
- The instantaneous velocity can be found by performing the derivation of x(t):
</span>
<span>So now we just have to substitute t=0, t=4 s and t=8 s:
- t=0: v(0)=0
- t=4 s: </span>
- t=8 s:
part c) <span>how long after starting from rest is the car again at rest?
- To solve this part we must find the value of t for which v(t)=0, so:
</span>
<span>The first solution is t=0 s, which corresponds to the beginning of the motion, so we are not interested in this value. The second solution is
</span>
<span>and this is the time at which the car is at rest again.</span>
Explanation:
It is given that, the force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of the car and the square of the car's speed such that,
mg is the weight of the car
r is the radius of the curve
v is the speed of the car
Case 1.
F = 640 pounds
Weight of the car, W = mg = 2600 pound
Radius of the curve, r = 650 ft
Speed of the car, v = 40 mph
k = 0.1
Case 2.
Radius of the curve, r = 750 ft
Speed of the car, v = 30 mph
F = 312 N
Hence, this is the required solution.