Question

Starting from position x0 = 0 at time t0 = 0, a bicyclist rides in a straight line a distance

x in time t and later is a distance 8x after a time 4t. What is the bicyclist’s constant
acceleration?
A: x/2t^2
B: 7x/3t
C: 6x/5t^2
D: 2x/3t^2

Answer 1

The constant acceleration of the bicyclist is found using the kinematic equation for uniformly accelerated motion. Starting from rest, at time t = 0, the acceleration is given by 2x/3t^2. This result matches option D.

Step-by-step explanation:

The subject of the question is Physics, specifically dealing with kinematics and the concept of constant acceleration. To find the constant acceleration of a bicyclist who travels a distance x in time t and then 8x in time 4t, starting from rest, we can use the kinematic equation for uniformly accelerated motion:

final position (x_f) = initial position (x_i) + initial velocity (v_i)t + (1/2)at^2

At time t, the final position is x, initial velocity is 0 (starting from rest), and initial position is 0.

x = 0*t + (1/2)*a*t^2
x = (1/2)*a*t^2
a = (2x)/t^2

At time 4t, the final position is 8x.

8x = 0*4t + (1/2)*a*(4t)^2
8x = 8*(1/2)*a*t^2
8x = 4*a*t^2
a = (2x)/t^2

So the constant acceleration a is equal to 2x/3t^2, which corresponds to option D.