U=10 m/s
v=30 m/s
t=6 sec
therefore, a=(v-u)/t
=(30-10)/6
=(10/3) ms^-2
now, displacement=ut+0.5*a*t^2
=60+ 0.5*(10/3)*36
=120 m
And you can solve it in another way:
v^2=u^2+2as
or, s=(v^2-u^2)/2a
=(900-100)/6.6666666.......
=120 m
Answer:
2.57 seconds
Explanation:
The motion of the ball on the two axis is;
x(t) = Vo Cos θt
y(t) = h + Vo sin θt - 1/2gt²
Where; h is the initial height from which the ball was thrown.
Vo is the initial speed of the ball, 22 m/s , θ is the angle, 35° and g is the gravitational acceleration, 9.81 m/s²
We want to find the time t at which y(t) = h
Therefore;
y(t) = h + Vo sin θt - 1/2gt²
Whose solutions are, t = 0, at the beginning of the motion, and
t = 2 Vo sinθ/g
= (2 × 22 × sin 35°)/9.81
= 2.57 seconds
Answer:
usually its because they dont wanna talk
Explanation:
I know what thats like. Either that or they don't trust you. It could be something personal or they just dont wanna talk. But i wouldn't push. Just let them talk if they want to
This is Kinematics and the equations in your book.
A speed time graph would plot the speed of something against the teime it was at a speed.
If it were changing it speed constantly, that would be a straight line if acclerating. Total distrance would be the area under the graph.
Answer:
Explanation:
Use the velocity formula to solve
In this question, you are given velocity 
, and you are given a distance, 
. Time in this question is what you'll need to find.
Start by rearranging the velocity formula, to isolate for t.
Start by multiplying both sides by t
Then divide both sides by v.
Now that you've isolated for time, sub in your values and calculate.
