Alexander's hobby is dirt biking. On one occasion last weekend, he accelerated from rest to 17.8 m/s in 1.56 seconds. He then maintained this speed for 9.47 seconds. Seeing a coyote cross the trail ahead of him, he abruptly stops in 2.79 seconds. Determine Alexander's average speed for this motion. Thanks =)
we know that avg speed = (total distance traveled) / (total time take).........(i) now total time = ( 1.56 + 9.47 + 2.79 ) secs. the total distance covered is calculated as follows: the whole journey can be divided into three parts: 1) when he accelerated from rest 2) when he moves with constant speed 3) when he decelerated and came to a stop. now, find the distance covered in each of these parts using the kinematic equations. 1) given: u1 =0, a1 = 17.8m/s^2, t1 = 1.56 sec s1 = u1t1 + 1/2 a1 t1^2 s1 = 0 + 1/2 (17.8) (1.56)^2 s1 = 8.9 (1.56)^2 m also, the constant speed acquired v1 = u1 + a1 t1 v1 = (17.8) (1.56) m/s 2) given: constant speed v1 = (17.8)(1.56) m/s, t2 = 9.47 sec s2 = v1 * t2 s2 = (17.8) (1.56) (9.47) m 3) given: v3 = 0, t3 = 2.79 s, v1 = u3 = (17.8)(1.56) m/s first find the deceleration [-a3] v3 = u3 - a3 t3 u3 = a3 t3 a3 = u3/ t3 a3 = (17.8) (1.56) / (2.79) m/s^2 (v3)^2 = (u3)^2 - 2 a3 s3 (u3)^2 = 2 a3 s3 s3 = (u3)^2 / 2 a3 s3 = (17.8)^2 (1.56)^2 (2.79) / 2 (17.8) (1.56) s3 = (17.8) (1.56) (2.79) / 2 s3 = 8.9 (1.56) (2.79) m so, total distance covered = s1 + s2 + s3. substitute this in (i) and u'll get the avg speed.
Sorry, there was an error in the question. He accelerates from 0 to 17.8 m/s, not to 17 m/s/s. Does the answer change with this error at all?
oh...yes. then the answer will change. sorry, i took it as acceleration. then the changes would be as follows: 1) given: u1 = 0, v1 = 17.8, t1 = 1.56 use v1 = u1 + a1 t1.......to find a1. then sub that s1 = 1/2 a1 t1^2......to get s1. 2) given: constant speed = v1 = 17.8, t2 = 9.47 calculate s2 using, s2 = v1 t2 3) here, proceed the original way. the only change will be, v1 = u3 = 17.8 now, u'll get the right answer.
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