Respuesta :
ANSWER:
If a ball is thrown into the air with a velocity of 34 feet per second, then velocity of the ball after 1 second is 2 feet per second
SOLUTION:
Given, a ball is thrown into the air with a velocity of 34 feet per second
Initial velocity (u) = 34 feet per second
And also given a relation between displacement and time = [tex]\mathrm{y}=34 \mathrm{t}-16 \mathrm{t}^{2}[/tex] --- eqn 1
We need to find the velocity when t = 1 ; v = ?
We know that, v = u + at and [tex]\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}[/tex]
where v is instantaneous velocity and u is initial velocity
a is acceleration
t is time interval
s is displacement
using the displacement and time relation eqn (1) we get
Now, when t = 1, displacement s = 34(1) – 16(1)
[tex]\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}=34-16[/tex]
[tex]34 \times 1+\frac{1}{2} \times a \times 1^{2}=18[/tex]
[tex]34+\frac{a}{2}=18[/tex]
[tex]\begin{array}{l}{\frac{a}{2}=18-34} \\\\ {\frac{a}{2}=-16} \\ {a=-16 \times 2} \\ {a=-32}\end{array}[/tex]
here, -ve sign indicates that object is in deceleration . so acceleration is -32 ft/s
now put a value in v = u + at
v = 34 + (-32)(1)
v = 34 – 32
v = 2 ft/s
Hence, velocity of the ball after 1 second is 2 ft/s
