Dear student,
write the equation of motion for each horse; in general you have
x(t)=x0+v0 t + 1/2 at^2
x(t)=position as a function of time
x0=initial position
v0=initial speed (which here is zero for both horses)
a=acceleration
t=time
for George, we have:
x(t)=0+0+1/2 (0.22)t^2
for Arthur, we have:
x(t)=70+0-1/2(0.3)t^2 (accel is negative since it is in the opposite direction of George's
they meet when the have the same value of x:
0.11t6^2=70-0.15t^2
0.26t^2=70 or t=16.4 s
plug this into the george equation to get:
x(t=16.4s)=0.11(16.4)^2=29.6 m
Eevee,
In an historical movie, two knights on horseback start from rest 79.0 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.319 m/s2, while Sir Alfred's has a magnitude of 0.391 m/s2. Relative to Sir George's starting point, where do the knights collide?
1 answer
1 answer