The results of the calculation are;
1) The acceleration is 0.2 m/s^2
2) The acceleration is 20 m/s^2
3) Can not be solved as the parameters are not complete
4) Has a final velocity of 114.7 m/s
5) The height is 4.9 m
6) The total time taken is 0.4 s
7) The final velocity is 190 m/s
The term acceleration is defined as the rate of change of the velocity with time. We shall now look at the solution to each of the problems.
1) Using;
v = u + at
v = final velocity
u = initial velocity
a = acceleration
t = time
a = v - u/t = 6 - 4/10
a = 0.2 m/s^2
2) v^2 = u^2 + 2as
s = distance
u = initial velocity
a = acceleration
v =final velocity
Given that it started from rest u = 0 m/s
v^2 = 2as
a = v^2/2s
a = (20)^2/ 2 * 10
a = 20 m/s^2
3) This problem can not be solved because the initial velocity is missing
4) v = u + gt
v = 100 + (9.8 * 1.5)
v = 114.7 m/s
5) h = ut + 1/2 gt^2
Since it was dropped from a height u = 0 m/s
h = 1/2 gt^2
h = 0.5 * 9.8 * (1)^2
h = 4.9 m
6) v = u + gt
At the maximum height v = 0 m/s
u = gt
t = u /g
t = 2/9.8
t = 0.2 s
To go up and come back 2(0.2) = 0.4 s
7) s = ut + 1/2 at^2
400 = (10 * 4) + 0.5 * (4)^2 * a
400 = 40 + 8a
400 - 40/8 = a
a = 45 m/s^2
Using;
v = u + at
v = 10 + 45 * 4
v = 190 m/s
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