Respuesta :
Answer:
Explanation:
Given
velocity at A is [tex]v=4\ ft/s[/tex]
For [tex]r=500\ ft[/tex]
velocity is increasing at [tex]\dot{v}=0.004t\ ft/s^2[/tex]
Tangential acceleration is given by
[tex]a_t=\frac{\mathrm{d} v}{\mathrm{d} t}[/tex]
[tex]a_t=0.004t=\frac{\mathrm{d} v}{\mathrm{d} t}[/tex]
[tex]\int 0.004tdt=\int dv[/tex]
[tex]\int dv=\int 0.004tdt[/tex]
[tex]v=0.002t^2+c[/tex]
at [tex]t=0\ v=4\ ft/s[/tex]
[tex]4=0.002\cdot 0+c[/tex]
[tex]c=4\ ft/s[/tex]
thus [tex]v=0.002t^2+4[/tex]
Velocity in terms of Displacement is given by
[tex]v=\frac{\mathrm{d} s}{\mathrm{d} t}[/tex]
[tex]\Rightarrow \int ds=\int \left ( 0.002t^2+4\right )dt[/tex]
[tex]\Rightarrow s=\frac{0.002t^3}{3}+4t[/tex]
When car has traveled [tex]\frac{3}{4}[/tex] th of distance i.e.
[tex]s=\frac{3}{4}\times (2\pi r)=\frac{3\pi r}{2}[/tex]
[tex]s=750\pi [/tex]
[tex]750\pi =\frac{0.002t^3}{3}+4t[/tex]
[tex]\Rightarrow \frac{0.002t^3}{3}+4t-2356.5=0[/tex]
on solving we get [tex]t=139.23\ s[/tex]
Thus velocity at [tex]t=139.23\ s[/tex]
[tex]v=42.76\ s[/tex]
(b)Acceleration when car has traveled three-fourth the way of track
normal acceleration [tex]a_n=\frac{v^2}{r}=\frac{(42.76)^2}{500}[/tex]
[tex]a_n=3.658\ m/s^2[/tex]
Tangential acceleration [tex]a_t at t=139.23\ s[/tex]
[tex]a_t=0.556\ m/s^2[/tex]
Net acceleration [tex]a_t=\sqrt{(a_n)^2+(a_t)^2}[/tex]
[tex]a_n=\sqrt{(3.658)^2+(0.556)^2}[/tex]
[tex]a_n=3.7\ m/s^2[/tex]
The magnitude of the velocity and acceleration when it has traveled three-fourths the way around the track are; 42.76 ft/s and a = 3.7 ft/s²
What is the magnitude of the acceleration?
We are given;
Radius; r = 500 ft
Velocity at A; v = 4 ft/s
Rate of increase of velocity; v' = 0.004t ft/s²
A) Tangential acceleration is gotten from the formula;
a_t = dv/dt
Thus;
0.004t = dv/dt
∫0.004t.dt = ∫dv
v = 0.002t² + c
Velocity at A is 4 ft/s which is where t = 0 s. Thus;
4 = 0.002(0)² + c
c = 4
v = 0.002t² + 4
formula for velocity with respect to change of distance with time is;
v = ds/dt
∫ds = ∫(0.002t² + 4)dt
s = ¹/₃0.002t³ + 4t
When the car has travelled ³/₄th of the distance, it means that;
s = ³/₄ × (2πr)
s = ³/₄ × (2π * 500)
s = 750π
Thus;
750π = ¹/₃0.002t³ + 4t
¹/₃0.002t³ + 4t - 750π = 0
Solving with online polynomial calculator gives t = 139.26 s and velocity at that time is v = 42.76 ft/s
B) Formula for centripetal acceleration is;
a_c = v²/r
a_c = 42.76²/500
a_c = 3.658 ft/s²
a_t = 0.004t = 0.004 * 139.26
a_t = 0.55704 ft/s²
Net acceleration is;
a = √(3.658² + 0.55704²)
a = 3.7 ft/s²
Read more about Magnitude of acceleration at; https://brainly.com/question/1597065
