As per the question the initial speed of the car [ u] is 42 m/s.
The car applied its brake and comes to rest after 5.5 second.
The final velocity [v] of the car will be zero.
From the equation of kinematics we know that
[ here a stands for acceleration]



Here a is taken negative as it the car is decelerating uniformly.
We are asked to calculate the stopping distance .
From equation of kinematics we know that
[here S is the distance]
![= 42*5.5 +\frac{1}{2} [-7.64] [5.5]^2 m](https://tex.z-dn.net/?f=%3D%2042%2A5.5%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5B-7.64%5D%20%5B5.5%5D%5E2%20m)
[ans]
Rt= ΣR = 40Ω
Vt= 80V
It= 80V/40Ω= 2A
V1= 15Ω*2A= 30V
V2= 20Ω*2A= 40V
V3= 5Ω*2A= 10V
time=distance/speed
1.6/100 secs = 0.016secs=16millisecs
Answer:
v = 6i + 12j + 4k
Explanation:
Find the magnitude of the direction vector.
√(3² + 6² + 2²) = 7
Normalize the direction vector.
3/7 i + 6/7 j + 2/7 k
Multiply by the magnitude of v.
v = 14 (3/7 i + 6/7 j + 2/7 k)
v = 6i + 12j + 4k
Formula for final velocity: Vf= vi+(a*t)
Vi- initial velocity, a=acceleration, t-time
Vf=vi+(at)
Vf= 0+(9.8m/s*2.8s)
Vf= 27.44 m/s
The acceleration of the Earth when dropping something would be 9.8 m/s
Here is an reference that can help you answer problems like these.
Hope this helps and good luck :)