Answer:
the velocity of the car is 0.875 m/s
Explanation:

therefore the V of car is 0.875 m
 
        
             
        
        
        
Answer:
v = 6t² + t + 2, s = 2t³ + ½ t² + 2t
59 m/s, 64.5 m
Explanation:
a = 12t + 1
v = ∫ a dt
v = 6t² + t + C
At t = 0, v = 2.
2 = 6(0)² + (0) + C
2 = C
Therefore, v = 6t² + t + 2.
s = ∫ v dt
s = 2t³ + ½ t² + 2t + C
At t = 0, s = 0.
0 = 2(0)³ + ½ (0)² + 2(0) + C
0 = C
Therefore, s = 2t³ + ½ t² + 2t.
At t = 3:
v = 6(3)² + (3) + 2 = 59
s = 2(3)³ + ½ (3)² + 2(3) = 64.5
 
        
             
        
        
        
Answer:
d. zero 
Explanation:
Constant velocity means the acceleration is zero. In this case the velocity does not change,
hope this helps you 
have a good day :)
 
        
             
        
        
        
Answer:
The acceleration of the car will be 
Explanation:
We have given that distance from stop sign s = 200 m 
Time t = 0.2 sec
We have to find the constant acceleration 
Now from second equation of motion 


So the acceleration of the car will be 