Question

plz. guys. answer......​

Answer 1

s(t) = t^3- 8t^2+19t-12

now, simply put the values of t...

(a) t=0s

s(0) = 0^3- 8*0^2 + 19*0 -12

s(0) = -12

the position of object is 12 metres on negative x axis.

(b) t=1s

s(1)= 1^3 -8*1^2 +19*1 -12

s(1)= 1-8+19-12

s(1)= 0

the position of the object is 0 metres, i.e., origin.

(c) t=2s

s(2)= 2^3 -8*2^2 +19*2 -12

s(2)= 8-32+38-12

s(2)= 2

the position of the object is 2 metres on positive x axis.

(d) t=3s

s(3)= 3^3 -8*3^2 +19*3 -12

s(3)= 27-72+57-12

s(3)= 0

the position of the object is again origin.

(e) displacement between t=1s to t=2s

displacement=| s(2)-s(1) |

=2 -0 metres

= 2 metres

(f) displacement= | s(3)-s(2) |

= |0-2| metres

=|-2| metres

=2 metres


to find the velocity divide the difference of s(t) by t. to find the acceleration , divide the difference of velocity by t.
for direction , if value of velocity/acceleration is positive then direction will be positive x axis and if value is negative, direction will be negative x axis.