Answer:
Explanation:
a )
While breaking initial velocity u = 62.5 mph
= 62.5 x 1760 x 3 / (60 x 60 ) ft /s
= 91.66 ft / s
distance trvelled s = 150 ft
v² = u² - 2as
0 = 91.66² - 2 a x 150
a = - 28 ft / s²
b ) While accelerating initial velocity u = 0
distance travelled s = .24 mi
time = 19.3 s
s = ut + 1/2 at²
s is distance travelled in time t with acceleration a ,
.24 = 0 + 1/2 a x 19.3²
a = .001288 mi/s²
= 2.06 m /s²
c )
If distance travelled s = .25 mi
final velocity v = ? a = .001288 mi / s²
v² = u² + 2as
= 0 + 2 x .001288 x .25
= .000644
v = .025 mi / s
= .0025 x 60 x 60 mi / h
= 91.35 mph .
d ) initial velocity u = 59 mph
= 86.53 ft / s
final velocity = 0
acceleration = - 28 ft /s²
v = u - at
0 = 86.53 - 28 t
t = 3 sec approx .
Answer:
The energy that the truck lose to air resistance per hour is 87.47MJ
Explanation:
To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

Our values are:




Replacing,


We need calculate now the energy lost through a time T, then,

But we know that d is equal to

Where
v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,

(per hour)
Therefore the energy that the truck lose to air resistance per hour is 87.47MJ
Answer:
The answer is B. hope this helps
Explanation: