Question

in an emergency braking exercise, a student driver stops a car travelling at 83 km/h w in a time of 4.0 s. what is the cars acceleration during this time? (the answer is expressed in units of $\text{m/s}^2$.)
5.8 w
21 e
5.8 e
21 w
-5.8 e
question 3 (1 point)
a toy car is moving at 13 cm/s when it begins accelerating at $1.4 \text{ cm/s}^2$. if the acceleration is uniform, what is the speed of the car after it has travelled a distance of 27 cm?
240 cm/s
93 cm/s
62 cm/s
16 cm/s

Explanation:

Step1: Convert initial speed to m/s

Initial speed $v_0 = 83\ \text{km/h} = 83 \times \frac{1000\ \text{m}}{3600\ \text{s}} \approx 23.06\ \text{m/s [W]}$

Step2: Define final speed and time

Final speed $v = 0\ \text{m/s}$, time $t=4.0\ \text{s}$

Step3: Calculate acceleration magnitude

Acceleration formula: $a = \frac{v - v_0}{t}$
$a = \frac{0 - 23.06}{4.0} \approx -5.8\ \text{m/s}^2$
The negative sign means acceleration is opposite to direction of motion (East, [E]).
---

Step1: List known values for toy car

Initial speed $u=13\ \text{cm/s}$, acceleration $a=1.4\ \text{cm/s}^2$, distance $s=27\ \text{cm}$

Step2: Use kinematic equation $v^2=u^2+2as$

$v^2 = 13^2 + 2\times1.4\times27$
$v^2 = 169 + 75.6 = 244.6$

Step3: Solve for final speed $v$

$v = \sqrt{244.6} \approx 16\ \text{cm/s}$

Answer:

1. 5.8 [E]
2. 16 cm/s