Answer:
See below ~
Explanation:
Part (a) :
We can say a body is in uniform acceleration if the acceleration of the object remains constant with respect to time throughout its motion.
Part (b) :
We can say a body is non-uniform acceleration if the acceleration of the body varies with respect to time throughout its motion.
C: It would follow a path perpendicular to the radius of its current orbit
When there is nothing to keep it going in circles, it will stop moving in circles around the sun and continue in a straight path. Think of it as a catapult. If the stick stops, the rock keeps flying straight at the speed and direction it was flung.
Answer
given,
v = (6 t - 3 t²) m/s
we know,


position of the particle

integrating both side

x = 3 t² - t³
Position of the particle at t= 3 s
x = 3 x 3² - 3³
x = 0 m
now, particle’s deceleration


a = 6 - 6 t
at t= 3 s
a = 6 - 6 x 3
a = -12 m/s²
distance traveled by the particle
x = 3 t² - t³
at t = 0 x = 0
t = 1 s , x = 3 (1)² - 1³ = 2 m
t = 2 s , x = 3(2)² - 2³ = 4 m
t = 3 s , x = 0 m
total distance traveled by the particle
D = distance in 0-1 s + distance in 1 -2 s + distance in 2 -3 s
D = 2 + 4 + 2 = 8 m
average speed of the particle



It tells us the number of protons that are present in the nucleus, the positively charged region of that atom.
The braking distance is given by 
Explanation:
When the driver of a car hits the pedal of the brakes, the car starts decelerating until it stops. Assuming the deceleration is constant, then the motion is a uniformly accelerated motion, so we can use the following suvat equation:

where
u is the initial speed of the car
v is the final speed of the car, which is zero because the car comes to rest:
v = 0
a is the acceleration of the car
s is the distance travelled by the car during the deceleration, so it is the braking distance
Therefore, re-arranging the equation for s, we find an expression for the braking distance:

Note that the sign of
is negative since the car is decelerating, therefore the final sign of
is positive.
Learn more about accelerated motion:
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