Arecord of travel along a straight path is as follows: (a) start from rest with constant acceleration of 2.62 m/s 2 for 12.8 s; (b) constant velocity of 33.536 m/s for the next 0.895 min; (c) constant negative acceleration of −11.2 m/s 2 for 5.44 s. what was the total displacement x for the complete trip? answer in units of m.

First it will accelerate with acceleration a = 2.62 m/s^2

now the distance that it will move with this acceleration is

d1 = 0.5 * 2.62 * 12.8^2 = 214.6 m

so it will cover first 241.6 m during acceleration

now it will cover next distance with constant speed

so the distance covered is given as

d2 = v* t

d2 = 33.536 * 0.895*60 = 1800.9 m

after this it will comes to rest by constant deceleration which is given as by a = -11.2 m/s^2

so the distance it will cover is given by

d3 = (v^2 - vi^2)/(2a)

d3 = (0^2 - 33.536^2) / (2*-11.2)

d3 = 50.2 m

so the total distance that it move is given by

d = d1 + d2 + d3

d = 214.6 + 1800.9 + 50.2

d = 2065.7

so the total displacement is 2065.7 m

15.3 cos 65 degrees trig

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