Question

(m/s)

Answer 1

a) The acceleration of the car is $0.33 m/s^2$

b) The new acceleration would be$0.17 m/s^2$

Explanation:

a)

The acceleration of a body is equal to the rate of change of the velocity:

$a=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the  change in velocity

$\Delta t$ is the time interval considered

For the data of the car in this problem, we have:

$v_1 = 0.13 m/s$ is the initial velocity

$v_2 = 0.36 m/s$ is the final velocity

$t_1 = 1.92 s$ is the initial time

$t_2 = 2.61 s$ is the final time

Substituting, we find the acceleration:

$a=\frac{v_2-v_1}{t_2-t_1}=\frac{0.36-0.13}{2.61-1.92}=0.33 m/s^2$

b)

To solve this part, we can apply Newton's second law of motion, which states that the net force applied on a body is equal to the product between its mass and its acceleration:

$F=ma$

where

F is the net force

m is the mass

a is the acceleration

The equation can be rewritten as

$a=\frac{F}{m}$

We observe that the acceleration is directly proportional to the force: therefore, if the force were cut in half, the acceleration will also halve. Therefore, the new acceleration would be

$a' = \frac{a}{2}=\frac{0.33}{2}=0.17 m/s^2$

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