Answer:
The transverse component of acceleration is 26.32
where as radial the component of acceleration is 8.77 
Explanation:
As per the given data
u=π/4 rad
ω=u'=2 rad/s
α=u''=4 rad/s

So the transverse component of acceleration are given as

Here


So

The transverse component of acceleration is 26.32 
The radial component is given as

Here

So

The radial component of acceleration is 8.77 
Here when car in front of us applied brakes then it is slowing down due to frictional force on it
So here we can say that friction force on the car front of our car is given as

So the acceleration of car due to friction is given as



now it is given that


so here we have


so the car will accelerate due to brakes by a = - 8.52 m/s^2
Answer:
1➡️ this is the method of decomposition
2➡️ H2 and O2
3➡️ b
sorry if I am wrong
We make use of the equation: v^2=v0^2+2a Δd. We substitute v^2 equals to zero since the final state is halting the truck. Hence we get the equation -<span>v0^2/2a = Δd. F = m a from the second law of motion. Rearranging, a = F/m
</span>F = μ Fn where the force to stop the truck is the force perpendicular or normal force multiplied by the static coefficient of friction. We substitute, -v0^2/2<span>μ Fn/m</span> = Δd. This is equal to