a ball is rolling down a hill at +1.0m/s the ball is uniformly accelerating at +4.9m/s^2 what will the balls velocity be 4.0 later

Respuesta :

we are given

initial velocity is +1.0 m/s

so, [tex]u=1.0m/s[/tex]

Let's assume final velocity as v

uniformly accelerating at +4.9m/s^2

so, [tex]a=4.9m/s^2[/tex]

time is 4 sec

so, t=4 s

now, we can use formula

[tex]a=\frac{v-u}{t}[/tex]

we can plug values

[tex]4.9=\frac{v-1}{4}[/tex]

[tex]4.9*4=4*\frac{v-1}{4}[/tex]

[tex]4.9*4=v-1[/tex]

[tex]19.6=v-1[/tex]

[tex]v=20.6m/s[/tex]...............Answer


Answer:

So velocity of ball after 4 sec will be 20.6 m/sec

Step-by-step explanation:

We have given initial velocity of the ball u = 1 m/sec

Acceleration of the ball [tex]a=4.9m/sec^2[/tex]

We have to find velocity after 4 sec

So ti,e t = 4 sec

According to first equation of motion we know that v = u+at

So [tex]v=1+4.9\times 4=20.6m/sec[/tex]

So velocity of ball after 4 sec will be 20.6 m/sec