A car starts off at a 0 mph, the final velocity is 203 mph and the acceleration is an average of 2.93 ft/s² and it is traveling up a 6.4% slope. If the straight line trail is 12.42 miles long, the time take the car to travel that distance is 0.075 hour = 270 s
Explanation:
Velocity is defined as the rate of change in displacement while acceleration is defined as the rate of change of velocity. Acceleration may be positive or negative. Acceleration is positive when the velocity of the object is increases and it is negative when velocity of the object is decreases. Negative acceleration is also called deceleration.
Mathematically
a =[tex]\frac{v_{f} - v_{i}}{t}[/tex]
Where a is the acceleration of the object, [tex]v_{f}[/tex] is the final velocity of the object and [tex]v_{i}[/tex] is the initial velocity of the object. t is equal to time taken.
Given data:
[tex]v_{f}[/tex] = 203 miles/h
[tex]v_{i}[/tex] = 0 miles/h
a = 2.93 ft/s²
d = 12.42 miles
t = ?
As velocities and acceleration given in different units, So we need to convert to obtain same units. Here we convert unit of acceleration from ft/s² to miles/h².
1 ft/s² = 2454.5454545455 miles/h²
2.93 ft/s² = 7191.8181818182 miles/h²
So,
a = 7191.81 miles/h²
Apply formula
a =[tex]\frac{v_{f} - v_{i}}{t}[/tex]
t = [tex]\frac{v_{f} - v_{i}}{a}[/tex]
t = 203/7191.81
t₁ = 0.028 hour = 101.61 s
We use 3rd equation of motion to calculate distance cover during this time interval
2ad = [tex]v_{f}^{2}[/tex] - [tex]v_{i}^{2}[/tex]
2*7191.81*d = 203 - 0
d = 2.86 miles
remaining distance = 12.42 - 2.86 = 9.56 miles
Now we use
d = v*t
t = d/v
t = 9.56/203
t₂ = 0.047 hour = 169.52 s
Total time = T = t₁ + t₂
T = 0.028 + 0.047
T = 0.075 hour = 270 s
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