Respuesta :
Answer:
The value of t₁ is 4.65 sec.
The value of t₂ is 85.35 sec.
The value of v is 60.45 m/s.
Explanation:
Given that,
Acceleration = 13.0 m/s²
Total distance [tex]d=5.30\times10^{3}\ m[/tex]
Total time T=90.0 sec
We need to calculate the distance
Using equation of motion
[tex]s_{1}=ut+\dfrac{1}{2}at^2[/tex]
Put the value into the formula
[tex]s_{1}=0+\dfrac{1}{2}\times13\times (t_{1})^2[/tex]....(I)
We need to calculate the second distance
Using formula of distance
[tex]s_{2}=vt[/tex]
Put the value into the formula
[tex]s_{2}=vt_{2}[/tex]....(II)
We need to calculate the velocity
Using equation of motion
[tex]v=ut+at[/tex]
Put the value into the formula
[tex]v=13t_{1}[/tex]....(III)
Put the value of v in equation (II)
[tex]s_{2}=13t_{1}t_{2}[/tex]
Now, on addition equation (I) and equation (III)
[tex]s_{1}+s_{2}=\dfrac{1}{2}\times13\times (t_{1})^2+13t_{1}t_{2}[/tex]
Put the value into the formula
[tex]5.30\times10^{3}=\dfrac{1}{2}\times13\times (t_{1})^2+13t_{1}t_{2}[/tex]
[tex]5.30\times10^{3}=\dfrac{1}{2}\times13\times (t_{1})^2+13t_{1}(90-t_{1})[/tex]
[tex]5.30\times10^{3}=\dfrac{1}{2}\times13\times (t_{1})^2+13t_{1}\times90-13(t_{1})^2[/tex]
[tex]6.5t_{1}^2-1170t_{1}+5.30\times10^{3}=0[/tex]
[tex]t_{1}=4.65\ sec[/tex]
Put the value in the equation (I)
[tex]s_{1}=0+\dfrac{1}{2}\times13\times (4.65)^2[/tex]
[tex]s_{1}=140.54\ m[/tex]
Now the value of [tex]t_{2}[/tex] is
[tex]t_{1}+t_{2}=90[/tex]
Put the value of t₁
[tex]t_{2}=90-4.65[/tex]
[tex]t_{2}=85.35\ sec[/tex]
The value of velocity is
Put the value in the equation (III)
[tex]v=13\times4.65[/tex]
[tex]v=60.45\ m/s[/tex]
The value of [tex]s_{2}[/tex]
Put the value of v and t₁ in equation (II)
[tex]s_{2}=vt_{1}[/tex]
[tex]s_{2}=60.45\times4.65[/tex]
[tex]s_{2}=281.1\ m[/tex]
Hence, The value of t₁ is 4.65 sec.
The value of t₂ is 85.35 sec.
The value of v is 60.45 m/s.
