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10. An object is released from a height.(a) Find its speed at (1) t= 1s, (2) t=2s, (3) t= 3s.(b) Find the distance traveled at (1) t=1s, (2) t=2s (3) t=3s |
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Answer» Answer: The body is at rest before it is being dropped from the tower. So the velocity of the body at this point is given by the EQUATION v = u + at where, v = final velocity, u = initial velocity, a = acceleration DUE to gravity and t = TIME taken. Therefore, when the body is at rest, t = 0, then v = 0 + 10(0) = 0 m. Now, the body is dropped from the tower, the final velocity is given by the equation, s=ut+ 2 1
at 2 . In this case, initial velocity u = 0, a = 10m/s 2 and the DISTANCE covered is calculated for every increase of 1 second. At t = 1s, s = 0 + 1/2(10)(1*1) = 5 m. At t = 2s, s = 0 + 1/2(10)(2*2) = 20 m. At t = 3s, s = 0 + 1/2(10)(3*3) = 45 m. At t = 4s, s = 0 + 1/2(10)(4*4) = 80 m. At t = 5s, s = 0 + 1/2(10)(5*5) = 125 m. At t = 6s, s = 0 + 1/2(10)(6*6) = 160 m. Therefore, the displacement 'd' every second 't' are as follows, t = 0, d = 0 m, t = 1, d = 5 m, t = 2, d = 20 m, t = 3, d = 45 m, t = 4, d = 80 m, t = 5, d = 125 m, t = 6, d = 160 m. |
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