Question

1. a toy car driving in a straight line produces the velocity vs time graph shown below. what is the cars displacement in meters between 0 and 6 seconds? a. 4 m b. 24 m c. 8 m d. 12 m e. 16 m 14. what is the acceleration of the car in question 13 between 4 and 6 seconds? a. 4 m/s² b. 2 m/s² c. 0 m/s² d. -2 m/s² e. -4 m/s² 15. a student has a mass on earth equal to \m\ and a weight equal to \w\. the student travels to a planet where the gravitational field strength is equal to 5 n/kg. on this planet what must be true about the students mass and weight? a. the students mass is less and the students weight is also less b. the students mass is less and the students weight is also less c. the students mass is the same and the students weight is the same d. the students mass is the same and the students weight is more e. the students mass is the same and the students weight is less 16. a student places a 1.4 kg object on flat ground. what is the magnitude of the normal force that acts on the object? a. 13.7 n b. 1.4 n c. 14.0 n d. 9.8 n e. 11.2 n

Explanation:

Question 13

Step1: Recall displacement - area under v - t graph

The area under a velocity - time graph gives the displacement. The graph is a trapezoid with bases $b_1 = 2$, $b_2=6$ and height $h = 4$.
The formula for the area of a trapezoid is $A=\frac{(b_1 + b_2)h}{2}$.

Step2: Calculate the area

Substitute $b_1 = 2$, $b_2 = 6$ and $h=4$ into the formula: $A=\frac{(2 + 6)\times4}{2}=\frac{8\times4}{2}=16$ m.

Step1: Recall acceleration formula

Acceleration $a=\frac{\Delta v}{\Delta t}$. The initial velocity $v_i = 4$ m/s at $t_i=4$ s and the final velocity $v_f = 0$ m/s at $t_f = 6$ s.

Step2: Calculate acceleration

$a=\frac{v_f - v_i}{t_f - t_i}=\frac{0 - 4}{6 - 4}=\frac{- 4}{2}=-2$ m/s².

Step1: Understand mass and weight concepts

Mass is a measure of the amount of matter in an object and is an inherent property, so it does not change with location. Weight $W=mg$, where $g$ is the gravitational field strength. On the new planet, $g = 5$ N/kg (less than Earth's $g\approx9.8$ N/kg), so the weight will be less.

Answer:

e. 16 m

Question 14