An object starts from rest and travels 20 m in first 2 seconds and 160 m in next 4 seconds . find its velocity after 7 seconds after the start.
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At A be U = 20 m/s
At B be V = 30 m/s
At C be W
For uniform acceleration motion, we know that
v² = u² + 2as
Now, here u = U, a = A, s = S, v = V.
Therefore, V² = U² + 2AS => S = (V² - U²)/2A—(1)
Now, for point M, s = S/2, u = U, v = W.
Therefore, W² = U² + 2A(S/2) => W² = U² + AS
Putting value of S in above equation from equation (1)—
W² = U² + A(V² - U²)/2A = U² + (V² - U²)/2
=> W² = (U² + V²)/2 => W =√(U²+V²)/2(U²+V²)/2
Now, put the values
W² = (20²+30²)/2 = (400+900)/2 = 1300/2 =650
W = √650
Approx 25.5 m/s
At B be V = 30 m/s
At C be W
For uniform acceleration motion, we know that
v² = u² + 2as
Now, here u = U, a = A, s = S, v = V.
Therefore, V² = U² + 2AS => S = (V² - U²)/2A—(1)
Now, for point M, s = S/2, u = U, v = W.
Therefore, W² = U² + 2A(S/2) => W² = U² + AS
Putting value of S in above equation from equation (1)—
W² = U² + A(V² - U²)/2A = U² + (V² - U²)/2
=> W² = (U² + V²)/2 => W =√(U²+V²)/2(U²+V²)/2
Now, put the values
W² = (20²+30²)/2 = (400+900)/2 = 1300/2 =650
W = √650
Approx 25.5 m/s
Similar questions
At B be V = 30 m/s
At C be W
For uniform acceleration motion, we know that
v² = u² + 2as
Now, here u = U, a = A, s = S, v = V.
Therefore, V² = U² + 2AS => S = (V² - U²)/2A—(1)
Now, for point M, s = S/2, u = U, v = W.
Therefore, W² = U² + 2A(S/2) => W² = U² + AS
Putting value of S in above equation from equation (1)—
W² = U² + A(V² - U²)/2A = U² + (V² - U²)/2
=> W² = (U² + V²)/2 => W =√(U²+V²)/2(U²+V²)/2
Now, put the values
W² = (20²+30²)/2 = (400+900)/2 = 1300/2 =650
W = √650
Approx 25.5 m/s