Question

A particle P moves in a straight line. The velocity vms−1 at time t s is given by
v = 5tt − 2 for 0 ≤ t ≤ 4,
v = k for 4 ≤ t ≤ 14,
v = 68 − 2t for 14 ≤ t ≤ 20,
where k is a constant.
(i) Find k. [1]
(ii) Sketch the velocity-time graph for 0 ≤ t ≤ 20. [3]
(iii) Find the set of values of t for which the acceleration of P is positive. [2]
(iv) Find the total distance travelled by P in the interval 0 ≤ t ≤ 20.

Answer 1

1) v = 5t² -2 0≤ t ≤ 4
so, at t = 4 sec
velocity ( U )= 5(4)² -2 = 78 m/s²

now, V = K ( constant ) so,
V = K = 78 m/s² in 4 ≤ t ≤ 14


2) see attachement

3) v = 5t² -2
differentiate wrt t
dv/dt = 10t at 0 ≤ t ≤ 4
hence, dv/dt ≥ 0 so, acceleration will be positive in interval [ 0 , 4]


(4) v = 5t² - 2
at t = 0 , v = -2
now,
dx/dt = 5t² -2

dx =5t²dt -2dt
integrate
x = (5/3)[ t³] - 2[ t ]
= 5× 64/3 -8 = (320-24)/3 = 296/3 m

again
v = K = 78
x = 78[ 14 -4 ] = 780 m

again ,
v = 68 -2t

dx = 68dt -2tdt = 68[ 20-14] -[ 20²-14²]
= 68× 6 - 34× 6
= 34 × 6 = 204 m

hence , total displacement = 296/3m + 780m + 204 m