We are given: Final velocity ()=20 m/s .
Time t= 2.51 s and
distance s = 82.9 m.
We know, equation of motion
Let us plug values of final velocity, and time in above equation.
Subtracting 2.51a from both sides, we get
-----------equation(1)
Using another equation of motion
Plugging values of vi =20-2.51a, t=2.51 and distnace s=82.9 in this equation.
We get,
Now, we need to solve it for a.
20-20+2.51a=165.8a.
-163.29a=0
a=0.
So, the acceleration would be 0 m/s^2.
Answer:
The speed at the end of the track = 27 m/s
The acceleration = 1.2 m/s²
Please find the Δx vs Δt, v vs Δt, a vs Δt
Explanation:
We have;
x = u·t + 1/2·a·t²
Where;
x = The distance = 300 m
u = The initial velocity = 0 m/s (Ball at rest)
t = The time taken = 22.4 s
Therefore;
300 = 0 + 1/2×a×22.4²
a = 2×300/22.4² = 1.19579 ≈ 1.2 m/s²
v = u + a×t
∴ v = 0 + 1.2 × 22.4 = 26.88 ≈ 27 m/s
Part of the table of values is as follows;
t, x, v
0, 0, 0
0.4, 0.095663, 0.478316
0.8, 0.382653, 0.956632
1.2, 0.860969, 1.434948
1.6, 1.530611, 1.913264
2, 2.39158, 2.39158
2.4, 3.443875, 2.869896
2.8, 4.687497, 3.348212
3.2, 6.122445, 3.826528
3.6, 7.748719, 4.304844
