Question

a car, starting from position of rest in X- direction moves with the equation x=9-2t+t², then after Three seconds the displacements will be_________m​

Answer 1

Answer:

12 m

Explanation:

substitue the value 3 in the equation

Answer 2

Answer:

Given,

Given,x=t

Given,x=t 2

Given,x=t 2 −4t+6

Given,x=t 2 −4t+6So, the change in velocity is:

Given,x=t 2 −4t+6So, the change in velocity is:dt

Given,x=t 2 −4t+6So, the change in velocity is:dtdx

Given,x=t 2 −4t+6So, the change in velocity is:dtdx

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2 =2

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2 =2So the magnitude of the distance traveled is:

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2 =2So the magnitude of the distance traveled is:6−2=4m

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2 =2So the magnitude of the distance traveled is:6−2=4mAt t=3,x

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2 =2So the magnitude of the distance traveled is:6−2=4mAt t=3,x 3

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2 =2So the magnitude of the distance traveled is:6−2=4mAt t=3,x 3

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2 =2So the magnitude of the distance traveled is:6−2=4mAt t=3,x 3 =3

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2 =2So the magnitude of the distance traveled is:6−2=4mAt t=3,x 3 =3So distance traveled from t=2s to t=3s

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2 =2So the magnitude of the distance traveled is:6−2=4mAt t=3,x 3 =3So distance traveled from t=2s to t=3sx

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2 =2So the magnitude of the distance traveled is:6−2=4mAt t=3,x 3 =3So distance traveled from t=2s to t=3sx 3

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2 =2So the magnitude of the distance traveled is:6−2=4mAt t=3,x 3 =3So distance traveled from t=2s to t=3sx 3

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2 =2So the magnitude of the distance traveled is:6−2=4mAt t=3,x 3 =3So distance traveled from t=2s to t=3sx 3 −x

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2 =2So the magnitude of the distance traveled is:6−2=4mAt t=3,x 3 =3So distance traveled from t=2s to t=3sx 3 −x 2

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2 =2So the magnitude of the distance traveled is:6−2=4mAt t=3,x 3 =3So distance traveled from t=2s to t=3sx 3 −x 2

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2 =2So the magnitude of the distance traveled is:6−2=4mAt t=3,x 3 =3So distance traveled from t=2s to t=3sx 3 −x 2 =3−2=1m

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2 =2So the magnitude of the distance traveled is:6−2=4mAt t=3,x 3 =3So distance traveled from t=2s to t=3sx 3 −x 2 =3−2=1mThus the total distance is:

Given,x=t 2 −4t+6So, the change in velocity is:dtdx =2t−4Since velocity is changing,At t=0,x 1 =6At t=2,x 2 =2So the magnitude of the distance traveled is:6−2=4mAt t=3,x 3 =3So distance traveled from t=2s to t=3sx 3 −x 2 =3−2=1mThus the total distance is:4+1=5m