<u>We are given:</u>
Initial velocity (u) = 32 m/s
Acceleration (a) = 3 m/s²
Displacement (s) = 40 m
Final Velocity (v) = v m/s
<u>Solving for the Final Velocity:</u>
from the third equation of motion:
v² - u² = 2as
<em>replacing the variables</em>
v² - (32)² = 2(3)(40)
v² = 240 + 1024
v² = 1264
v = √1264
v = 35.5 m/s
Therefore, the velocity of the bike after travelling 40 m is 35.5 m/s
Answer:
acceleration of the car is 3 m\s^2
Explanation:
from rest means the initial velocity (vi) is zero
time = 5s
final velocity (vf) = 15m\s
a = vf - vi \ t
a = (15-0) \ 5
a= 3 m\s^2
which means that the car is speeding up 3 meters every second
Answer:
x = 7226.94 m
y = 1677.4 m
Explanation:
This is a classic problem of a parabolic move. Let's analyze the given data.
The speed is 300 m/s but it's fired at 55°, so the x and y component of the speed would be:
Vx = 300 cos55 = 172.07 m/s
Vy = 300 sin55 = 245.75 m/s
Now, to get the x and y components of the shell, we need to apply the following expressions:
X = Xo + Vx*t (1)
Y = Yo + Vy*t - 1/2 gt² (2)
Xo and Yo is 0, and we already have the speed in x and y, so the components are:
X = 0 + (172.07)(42)
<h2>
X = 7226.94 m</h2>
For the y component:
Y = 0 + (245.75)(42) - 1/2 (9.8)(42)²
Y = 10321.5 - 8643.6
<h2>
Y = 1677.4 m</h2><h2>
</h2>
Hope this helps
