Question

1. if l represents the dimensions of length and t that of time, then the dimensions of acceleration are (a) l - t (b) l/t (c) t/l (d) l/t (e) none of these
2. in the absence of air friction, an object dropped near the surface of the earth experiences a constant acceleration of about 9.8 m/s². this means that the (a) speed of the object increases 9.8 m/s during each second (b) speed of the object as it falls is 9.8 m/s (c) object falls 9.8 meters during each second (d) object falls 9.8 meters during the first second only (e) derivative of the distance with respect to time for the object equals 9.8 m/s²

the graph above shows the velocity v as a function of time t for an object moving in a straight line. which of the following graphs shows the corresponding displacement x as a function of time t for the same time interval?
questions 4 - 5
at time t = 0, car x traveling with speed v₀ passes car y, which is just starting to move. both cars then travel on two parallel lanes of the same straight road. the graphs of speed v versus time t for both cars are shown above.

1. which of the following is true at time t = 20 seconds? (a) car y is behind car x. (b) car y is passing car x. (c) car y is in front of car x. (d) both cars have the same acceleration. (e) car x is accelerating faster then car y.
2. from time t = 0 to time t = 40 seconds, the areas under both curves are equal. therefore, which of the following is true at time t = 40 seconds? (a) car y is behind car x. (b) car y is passing car x. (c) car y is in front of car x. (d) both cars have the same acceleration. (e) car x is accelerating faster than car y.

Explanation:

Question 1

Step1: Recall acceleration formula

Acceleration $a=\frac{\Delta v}{\Delta t}$, where $\Delta v$ is change in velocity (has dimension of length per time: $[L/T]$) and $\Delta t$ has dimension of time $[T]$.

Step2: Calculate dimension

$a=\frac{[L/T]}{[T]}=\frac{[L]}{[T]^2}=[L/T^{2}]$. But among the given options, acceleration is defined as change in velocity per - time, and velocity has dimension of length per time. So acceleration has dimension of length per time squared which can also be thought of as $[L/T^{2}]$ and the closest in the given options is $[L/T^{2}]$ (equivalent to $[L/T\cdot T]$). The correct dimension of acceleration in terms of length $[L]$ and time $[T]$ is $[L/T^{2}]$ and the closest option is (B) $[L/T^{2}]$ (assuming a mis - typing in the options where (B) is meant to be $[L/T^{2}]$).

Acceleration is the rate of change of velocity. An acceleration of $9.8\ m/s^{2}$ means that the velocity (speed in the case of one - dimensional motion without direction change considerations here) of the object increases by $9.8\ m/s$ every second.

The velocity - time graph's slope gives acceleration and the area under the velocity - time graph gives displacement. When velocity is constant, displacement increases linearly with time. When velocity is zero, displacement does not change (horizontal line in displacement - time graph). When velocity is negative, displacement decreases. The area under the given velocity - time graph implies that displacement first increases linearly (constant positive velocity), then remains constant (zero velocity), and then decreases linearly (constant negative velocity).

Answer:

B. $[L/T^{2}]$

Question 2