Answer:
a) 4.40 s
b) 2.20 s
Explanation:
Given parameters are:
At constant power ,
initial speed of the car, [tex]v_0=0[/tex]
final speed of the car, [tex]v=32[/tex] mph
At full power,
initial speed of the car, [tex]v_0=0[/tex]
final speed of the car, [tex]v=64[/tex] mph
a)
At constant power, [tex]KE = \frac{1}{2} mv^2[/tex]
At full power, [tex]KE = \frac{1}{2} m(2v)^2[/tex]
So [tex]KE_f = 4KE_i[/tex]
So, time to reach 64 mph speed is 4 times more than the initial time
[tex]t = 4*1.10 =4.40[/tex] s
b)
[tex]v=v_0+at\\a=\frac{v-v_0}{t}=\frac{32-0}{1.1/3600}=104727.27[/tex] [tex]miles/hours^2[/tex]
For final 64 mph speed,
[tex]v=v_0+at\\t=\frac{v-v_0}{a}=\frac{64-0}{104727.27} = 6.111*10^{-4}[/tex] [tex]hours[/tex] = [tex]6.111*10^{-4}*3600=2.20[/tex] s
