Answer:
√(6ax)
Explanation:
Hi!
The question states that during a time t the motorcyle underwent a displacement x at constant acceleration a starting from rest, mathematically we can express it as:
x=(1/2)at^2
Then the we need to find the time t' for which the displacement is 3x
3x=(1/2)a(t')^2
Solving for t':
t'=√(6x/a)
Now, the velocity of the motorcycle as a function of time is:
v(t)=a*t
Evaluating at t=t'
v(t')=a*√(6x/a)=√(6*x*a)
Which is the final velocity
Have a nice day!
Answer:
because it can be hard
Explanation:
I said that because they be on bed rest
Answer:
0.8712 m/s²
Explanation:
We are given;
Velocity of first car; v1 = 33 m/s
Distance; d = 2.5 km = 2500 m
Acceleration of first car; a1 = 0 m/s² (constant acceleration)
Velocity of second car; v2 = 0 m/s (since the second car starts from rest)
From Newton's equation of motion, we know that;
d = ut + ½at²
Thus,for first car, we have;
d = v1•t + ½(a1)t²
Plugging in the relevant values, we have;
d = 33t + 0
d = 33t
For second car, we have;
d = v2•t + ½(a2)•t²
Plugging in the relevant values, we have;
d = 0 + ½(a2)t²
d = ½(a2)t²
Since they meet at the next exit, then;
33t = ½(a2)t²
simplifying to get;
33 = ½(a2)t
Now, we also know that;
t = distance/speed = d/v1 = 2500/33
Thus;
33 = ½ × (a2) × (2500/33)
Rearranging, we have;
a2 = (33 × 33 × 2)/2500
a2 = 0.8712 m/s²
