Answer:

Negative sign shows that velocity of the car is decreases at a constant rate
Explanation:
We have given velocity of the car is decreases from 32 m /sec to 24 m/sec in 4 second
So initial velocity of the car u = 32 m /sec
And finally car reaches to a velocity of 24 m/sec
Time taken to change in velocity = 4 sec
So final velocity v = 24 m/sec
From first equation of motion v = u+at
So 

Negative sign shows that velocity of the car is decreases at a constant rate
Answer:
15 because 5×5×5 is the same thing as 5×3 which equals to 15
Units of impulse: N • s, kg • meters per second
Explanation:
Impulse is defined in two ways:
1)
Impulse is defined as the product between the force exerted in a collision and the duration of the collision:

where
F is the force
is the time interval
Since the force is measured in Newtons (N) and the time is measured in seconds (s), the units for the impulse are
![[I] = [N][s]](https://tex.z-dn.net/?f=%5BI%5D%20%3D%20%5BN%5D%5Bs%5D)
So,
N • s
2)
Impulse is also defined as the change in momentum experienced by an object:

where the change in momentum is given by

where m is the mass and
is the change in velocity.
The mass is measured in kilograms (kg) while the change in velocity is measured in metres per second (m/s), therefore the units for impulse are
![[I]=[kg][m/s]](https://tex.z-dn.net/?f=%5BI%5D%3D%5Bkg%5D%5Bm%2Fs%5D)
so,
kg • meters per second
Learn more about impulse:
brainly.com/question/9484203
#LearnwithBrainly
Answer:
Velocity of truck will be 20.287 m /sec
Explanation:
We have given mass of the truck m = 4000 kg
Radius of the turn r = 70 m
Coefficient of friction 
Centripetal force is given 
And frictional force is equal to 
For body to be move these two forces must be equal
So 

Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,

Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?

a = -1.47 m/
(a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/
(since same friction force is applied)

s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m