Final answer:
To find the velocity of the heart movement from 1 cm to -6 cm over 2 seconds, the velocity is calculated to be -3.5 cm/s. Graphs of d-t, x-t, v-t, and a-t would show a straight line with a negative slope, a horizontal line at -3.5 cm/s for velocity, and a horizontal line at 0 m/s² for acceleration respectively.
Step-by-step explanation:
To calculate the velocity of a heart moving from 1 cm to -6 cm over 2 seconds, we use the formula v = d/t, where v is velocity, d is displacement, and t is time. The displacement is the final position minus the initial position, so in this case, d = (-6 cm) - (1 cm) = -7 cm. The time is 2 seconds, therefore, velocity v = (-7 cm)/2 s = -3.5 cm/s. The negative sign indicates the direction is opposite to the positive direction defined in the problem statement.
To draw the requested graphs, consider the following descriptions:
- d-t (distance vs. time): Since the problem implies a uniform velocity, the graph would be a straight line with a negative slope from 1 cm at t=0 to -6 cm at t=2 s.
- x-t (position vs. time): This would be identical to the d-t graph for uniform motion since position and displacement are the same in this context.
- v-t (velocity vs. time): This graph would be a horizontal line at v = -3.5 cm/s, indicating constant velocity.
- a-t (acceleration vs. time): As the velocity is constant, acceleration is zero, so the graph would be a horizontal line at a = 0 m/s².