Respuesta :
Explanation:
It is given that,
The velocity of a particle is given by :
[tex]v=20t^2-100t+50[/tex]
Where
v is in m/s and t is in seconds
Let a is the acceleration of the object at time t. So,
[tex]a=\dfrac{dv}{dt}[/tex]
[tex]a=\dfrac{d(20t^2-100t+50)}{dt}[/tex]
[tex]a=40t-100[/tex]
When a = 0
[tex]40t-100=0[/tex]
t = 2.5 s
a is zero at t = 2.5 s. Velocity, [tex]v=20(2.5)^2-100(2.5)+50[/tex]
v = -75 m/s
Since, [tex]v=\dfrac{ds}{dt}[/tex], s is the distance travelled
[tex]s=\int\limits{vdt}[/tex]
[tex]s=\int\limits{(20t^2-100t+50)dt}[/tex]
[tex]s=\dfrac{20t^3}{3}-50t^2+50t[/tex]
At t = 2.5 s, [tex]s=\dfrac{20(2.5)^3}{3}-50(2.5)^2+50(2.5)[/tex]
s = −83.34 m
Hence, this is the required solution.
Answer:
velocity = 50m/s
distance = -83.33m
Explanation:
The velocity of the particle is given by;
v = 20t² - 100t + 50 -------------------(i)
Since acceleration is the time rate of change in velocity, to get the acceleration (a), we find the derivative of equation (i) with respect to t as follows;
a = [tex]\frac{dv}{dt}[/tex] = [tex]\frac{d(20t^{2} - 100t + 50)}{dt}[/tex]
a = 40t - 100 ------------------(ii)
Now, when a = 0, let's find the time t by substituting the value of a into equation (ii) as follows;
0 = 40t - 100
=> 40t = 100
=> t = [tex]\frac{100}{40}[/tex]
=> t = 2.5seconds.
This means that at t = 2.5 seconds, the acceleration of the particle is zero(0)
(a) Now, to get the velocity at this instant (t = 2.5s), substitute the value of t into equation (i) as follows;
v = 20(0)² - 100(0) + 50
v = 0 - 0 + 50
v = 50 m/s
Therefore, the velocity when a is zero is 50m/s
(b) To get the distance (s) travelled at that instant, we integrate equation (i) as follows;
s = [tex]\int\limits^a_b {v} \, dt[/tex] -----------------------(iii)
Where;
a = the time instant = 2.5 seconds
b = the initial time instant = 0
v = 20t² - 100t + 50
Substitute these values into equation (iii) as follows;
s = [tex]\int\limits^a_b {(20t^{2} -100t + 50)} \, dt[/tex]
s = [tex]\frac{20t^{3} }{3}[/tex] - [tex]\frac{100t^{2} }{2}[/tex] + 50t [tex]|^{a}_{b}[/tex]
Substitute the values of a and b as follows;
s = [[tex]\frac{20(2.5)^{3} }{3}[/tex] - [tex]\frac{100(2.5)^{2} }{2}[/tex] + 50(2.5)] - [[tex]\frac{20(0)^{3} }{3}[/tex] - [tex]\frac{100(0)^{2} }{2}[/tex] + 50(0)]
s = [[tex]\frac{20(2.5)^{3} }{3}[/tex] - [tex]\frac{100(2.5)^{2} }{2}[/tex] + 50(2.5)] - 0
s = 104.17 - 312.5 + 125
s = -83.33m
Therefore, the distance traveled at that instant is -83.33m
