Question

The position coordinate of a particle which is confined to move along a straight line is given by
S = 2t*3- 24t+6, where s is measured in meters from a convenient origin and t is in seconds. Determine
(a) the time required for the particle to reach a velocity of 72 m/s from its initial condition at t = 0, (b) the
acceleration of the particle when v = 30 m/s, and (c) the net displacement of the particle during the
interval from t = 1 s to t = 4 s.​

Answer 1

Answer:

a)  4 seconds

b) 7.8 m/s$\frac{m}{sec^2}$

c) 64 metres

Explanation:

∵ S = 2t × 3 - 24t + 6

S = 6t - 24t + 6

a) v = 72 m/sec

at t=0;

s=6 m

∵ v=u+at

72=0+at

at=72

Now,

∵ s= ut + $\frac{1}{2} at^{2}$

6=0+$\frac{1}{2} at^{2}$

12=$at^{2}$

∴ at × t = 12

72 × t=12

t=$\frac{1}{6}$  seconds.

b) ∵ v = 30 m/s

v = u+at

30 = 0+a × $\frac{1}{6}$

a=180$\frac{m}{sec^{2} }$.

c) at t=1sec

s= 6 × 1-24 × 1+6

s= -12 m.

at t=4sec

s= 6 × 4-24 × 4+6

s= -76 m.

∴ The total displacement = 64 m.