Question

1. A particle moves in the one direction, stops, and then heads back in the opposite direction The position of a particle (in meters) as a function of time is given by x(t)- -3.65 t2 +4.52 t+ 7.91. The position of the particle where it stops is (a) 0 m. (b) 9.31 m. (c) 11.7 m. (d) -11.7 m. (e)-9.31 m. 2. If you drop an object from the top of a high building, how long does it take for the object to reach a speed of 100 km/h? (a) 0.124 s (b) 2.83 s (c) 5.11 s (d) 9.81 s (e) 12.3 s

Answer 1

(a)

the given equation

x(t) = -3.65 t^2 + 4.52 t + 4.91

differentiate above equation with respect to t

d ( x(t)/ dt = v

d/ dt ( -3.65t^2 +4.52t +7.91) = v

v= -7.3 t +4.52

when particles stop v = 0

0=-7.3t +4.52

t = 0.6191 s

x(t) = -3.65(0.6191)^2 +4.52(0.6191) +7.91

=9.31 m

option (b) is correct answer

(2)

apply kinematic equation

v= u+ at

t = v-u/a

= 100 ( 5/18) -0/9.8 m/s^2

=2.83 s

option (b) is correct answer