Answer to Question #80809 in Mechanics | Relativity for Desmond

Question

A car travelling with constant acceleration has a velocity of 6 m/s at a given instant. In the next minute it travels 3 km. Find its acceleration and the velocity after the minute.

Answer 1

Denote the initial velocity V_0=6 m⁄s, time motion T=1 min=60 s, and traveled path S=X-X_0=3 km=3,000 m, where X_0=X(0) - initial position of the car and X=X(T) final position of the car, at the moment T= 1min. a - acceleration
Let compose the kinematic equation of motion:
(X(t)=X_0+V_0⋅t+(a⋅t^2)/2@V(t)=V_0+a⋅t)┤
t=T=1 min⁡〖=60 sec〗
S=X(T)-X_0=3 km
V_0=V(0)=6 m⁄s
(a⋅T^2)/2=S-V_0⋅T
a=2⋅((S-V_0⋅T))/T^2
a=2⋅((3000-6⋅60))/〖60〗^2 =2⋅(3000-360)/3600=(2⋅2640)/3600=5280/3600=1.4666… m⁄s^2 =1 7/15 m⁄s^2
Then
V(T)=V_0+a⋅T=V_0+2 ((S-V_0⋅T))/T^2 ⋅t=V_0+2 ((S-V_0⋅T))/T
V(T)=6+2⋅((3000-6⋅60))/60=6+2⋅(3000-360)/60=6+(2⋅2640)/60=6+5280/60=6+88=
=94 m⁄s