Answer:
a.[tex]10.2mm/s^2[/tex]
b.25.2 s
Explanation:
We are given that
Initial velocity=0
Tangential acceleration=[tex]a_t=10mm/s^2[/tex]
[tex]r=0.8 m=0.8\times 10^3 mm=800mm[/tex]
By using [tex]1 m=10^3 mm[/tex]
a.t=4 s
[tex]a_t=\frac{v-u}{t}[/tex]
Using the formula
[tex]10=\frac{v}{4}[/tex]
[tex]v=4\times 10=40mm/s[/tex]
Normal acceleration=[tex]a_N=\frac{v^2}{r}[/tex]
Substitute the values
Normal acceleration=[tex]a_N=\frac{(40)^2}{800}=2mm/s^2[/tex]
Magnitude of acceleration=a=[tex]\sqrt{a^2_t+a^2_N}[/tex]
Using the formula
Magnitude of acceleration=a[tex]=\sqrt{(10)^2+(2)^2}=10.2mm/s^2[/tex]
Magnitude of acceleration=a=[tex]10.2mm/s^2[/tex]
b.Magnitude of acceleration=a=[tex]80mm/s^2[/tex]
[tex]a^2=a^2_t+a^2_N[/tex]
Substitute the values
[tex](80)^2=(10)^2+a^2_N[/tex]
[tex]6400=100+a^2_N[/tex]
[tex]a^2_N=6400-100=6300[/tex]
[tex]a_N=\frac{a^2_t t^2}{r}[/tex]
[tex]\sqrt{6300}=\frac{(10)^2t^2}{800}[/tex]
[tex]t^2=\frac{\sqrt{6300}\times 800}{100}[/tex]
[tex]t=\sqrt{\frac{\sqrt{6300}\times 800}{100}}[/tex]
[tex]t=25.2 s[/tex]
Hence, the time for the magnitude of the acceleration to be [tex]80mm/s^2[/tex]=25.2 s
The magnitude ol the acceleration when t : 4s is 10.2mm/s^2 and the time for the magnitude of the acceleration to be 80 mm/s^2 is 25.2s.
The initial velocity is zero
Tangential acceleration a1= 10mm/s^2
r= 0.8 x 10^3 mm= 800mm
Using the formula
10= v/4
v= 4 x 10= 40mm/s^2
Magnitude of acceleration
a= 80 mm/s^2
a^2= a^2t a^2n
Putting in the values
t= 25.2s
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