The velocity (in m/s) of a particle which moves along the x-axis is given by vx
ID: 2995703 • Letter: T
Question
The velocity (in m/s) of a particle which moves along the x-axis is given by vx = 2t2 + 14t -10, where the time is quoted in seconds. Determine (a) The distance from the origin travelled in 4 seconds (b) The acceleration of the particle after 4 seconds If an additional movement had been introduced in the y-axis at t = 0, where the particle starts from rest (in the y-axis) and accelerates at ay = 31, (Note - this movement is in addition to the velocity in the x-axis as stated above) (c) What is the total magnitude and direction of the velocity after 5 seconds?Explanation / Answer
(a) distance = integral (vdt) from 0 to 4s
= { 2t^3/3 + 7t^2 - 10t ) from 0 to 4
= 114.66 m
(b) acceleration = dv/dt = 4t +14
at t= 4s => a = 4*4 +14 = 30m/s^2
(c) ay = dvy/dt = 3t
vy = 3t^2/2
at t= 5 s vy = 3*5^2/2 = 37.5 m/s
vx = 110 m/s
net velocity = sqrt(vx^2 +vy^2 ) = 116.2164 m/s
direction = arctan(vy/vx) = 18.8247 degrees from poitive x axis in anticlockwise direction
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