Question

while moving at 40 m/s, an object accelerates uniformly by 6 m/s squared in 9 sec.
a. what is the velocity after this time interval?
b. how far does it go during this period?

Answer 1

hi
here is your solution
u= 40 m/s
a=6m/s²
t=9sec
v=?
s=?

using first equation of motion
v=u+at
v=40+6(9)
v=40+54
v=94m/s

using third equation of motion
v²-u²=2as
94²-40²=2(6)s
8836-1600=12s
7236=12s
s=7236/12
s=603m

hope it helps

Answer 2

Hii..
Here is your answer..

Solution:-
Initial velocity(u)=40m/s
Acceleration(a)= 6m/s^2
Time= 9 Sec

We'll find final velocity by using equation:-
$= > v = u + at \\ = > v = 40 + 6 \times 9 \\ = > v = 40 + 54 \\ = > v = 94$
Therefore,
Final velocity=94m/s

Now,
We have to find the distance travelled by him, we will find it by using equation:-
$= > s = ut + \frac{1}{2} a {t}^{2}$
$= > s = 40 \times 9 + \frac{1}{2} \times 6 \times {(9)}^{2} \\ = > s = 360 + \frac{1}{2} \times 6 \times 81 \\ = > s = 360 + 243$
$= > s = 603m$
Hope it helps uh...✌️✌️✌️