Answer:
6.746 ft/s^2
Explanation:
v(t)=50
v(0)=27
t=5/3600 = 1/720 hours
v(t)-v(0)= a(t-0)
50-27= a(1/720)
a= 23*720= 16560 mi/h^2
16560mi/h^2 * 5280/3600^2 (ft/s^2) =6.746 ft/s^2
The acceleration required to increase the speed of a car from 27 mi/h to 50 mi/h in 5 seconds is 6.75 ft/s².
Acceleration is given by:
[tex] a = \frac{\Delta v}{\Delta t} = \frac{v_{f} - v_{i}}{t_{f} - t_{i}} [/tex]
Where:
[tex] v_{f} [/tex]: is the final speed of the car = 50 mi/h
[tex] v_{i}[/tex]: is the initial speed of the car = 27 mi/h
[tex] \Delta t[/tex]: is the change in time = 5 seconds
The constant acceleration required is:
[tex] a = \frac{v_{f} - v_{i}}{\Delta t} [/tex]
[tex] a = \frac{50 \frac{mi}{h}*\frac{1 h}{3600 s}*\frac{5280 ft}{1 mi} - 27 \frac{mi}{h}*\frac{1 h}{3600 s}*\frac{5280 ft}{1 mi}}{5 s} [/tex]
[tex] a = 6.75 ft/s^{2} [/tex]
Therefore, the constant acceleration required is 6.75 ft/s².
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