Answer:
B). 3.4 s
Explanation:
As we can see the graph is given between velocity and time
so here we can see that the velocity is changing here with time and initially for some time it moves with constant speed
Then it's speed decreases to next few second and then speed increases to its maximum value
The time after which velocity comes to its maximum value will reach after t = 3 s
so out of the all given options most correct option will be

Answer:
-0.8 m/s²
Explanation:
Acceleration is the slope of a velocity vs. time graph.
a = Δv / Δt
a = (0 m/s − 12 m/s) / (15 s − 0 s)
a = -0.8 m/s²
Answer
Assuming
east is the positive x direction
north is the positive y direction
initial velocity , u = 19 j m/s
a)
acceleration , a = 1.6 j m/s^2
Using first equation of motion
v = u + a × t
v = 19 + 5.6× 1.6
v = 28 j m/s
the velocity of the car after 5.6 s is 28 m/s north
b)
acceleration , a = -1.5 j m/s^2
Using first equation of motion
v = u + a × t
v = 19 - 5.6 ×1.5
v = 10.6 j m/s
the velocity of the car after 5.6 s is 10.6 m/s north
Saying no and not throwing fits and manners.