Question

Just this last one!!

Answer 1

Answer:

$47 \ \frac{m}{s}$

Explanation:

s = displacement (m)

u = initial velocity $(\frac{m}{s})$

v = final velocity $(\frac{m}{s})$

a = acceleration $(\frac{m}{s^{2} })$

t = time (s)

s = 235

a = -4.7

v = 0

v² = u² + 2as

(0)² = u² + 2(-4.7)(235)

u² - 2209 = 0

u² = 2209

u = 47

Answer 2

Answer:

$\boxed {\boxed {\sf 47 \ m/s}}$

Explanation:

We are asked to find the initial velocity of the car before it began to accelerate.

We are given the acceleration, distance, and final velocity, so we will use the following kinematic equation:

${v_f}^2 = {v_i}^2 + 2ad$

The car's acceleration is -4.7 meters per second square. It traveled a distance of 235 meters. It came to rest, or a final velocity of 0 meters per second.

• a= -4.7 m/s²
• d= 235 m
• $v_f$= 0 m/s

Substitute the values into the formula.

$(0 \ m/s)^2 = {v_i}^2 + 2 (-4.7 \ m/s^2)(235 \ m)$

$0 = {v_i}^2 + 2 (-4.7 \ m/s^2)(235 \ m)$

Multiply the numbers in parentheses.

$0= {v_i}^2 + (-2209 \ m^2 / s^2)$

Add -2209 to both sides of the equation.

$0+ 2209 \ m^2 /s^2 = {v_i}^2+ ( -2209 \ m^2 /s^2 )+ 2209 \ m^2 /s^2$

$2209 \ m^2 /s^2 = {v_i}^2$

Take the square root of both sides.

$\sqrt {2209 \ m^2 /s^2} = \sqrt {{v_i}^2$

$\sqrt {2209 \ m^2 /s^2} = v_i$

$47 \ m/s = v_i$

The inital velocity of the car was 47 meters per second.