You used three methods of analysis to find the acceleration of the cart on the track: a two-point difference from v(t), an average of a(t) that was itself calculated by a succession of first differences of the v(t) data, and a line fit to v(t). some criteria to you consider which method is best are part a in which type of graph(s) is it easiest to see whether the acceleration was constant?

The correct method to obtain the constant acceleration of the cart is to obtain a straight line of the velocity-time data on the graph.

Explanation:

The acceleration of an object is explained as change in velocity of the body every second while it is in motion. If the rate at which the velocity of object changes remains constant, acceleration of object remains constant.

The acceleration is basically the magnitude of increment in velocity of the body after every second. The graph between the velocity and time is the plot of the velocity of body at different instants.

The slope of the velocity time graph of the body is the rate of change of velocity of the body. The slope of the v-t graph keeps changing, then the velocity of the body changes with a non-constant acceleration.

Whereas if the slope of a body is a straight line, then the acceleration of the body is constant and its velocity also changes at a constant rate.

Thus, the correct method to obtain the constant acceleration of the cart is to obtain a straight line of the velocity-time data on the graph.

Learn More:

1. Change in momentum of the car due to collision

2. The acceleration of the book kept on the table

3. Magnitude of acceleration of the car

Answer Details:

Grade: High School

Subject: Physics

Chapter: Acceleration

Keywords:

acceleration, rate of change, velocity, velocity-time graph, constant, non-constant, slope, straight line, cart, three methods.

answer: a. same unit

explanation:

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s