Question

The braking distance of a car moving at a velocity of v1 = 20m/s is equal to 40m. What will the braking distance S2 be at velocity of v2=25 m/s if it undergoes the same acceleration in both cases? 75.5 70 62.5 44.5 55 решение

Answer 1

Ответ:

Data:

S_1 = 40 m

V_1 = 20 m/s

V_2 = 25 m/s

V_x = 0 m/s

To Find:

a, S_2

Answer:

S = (V²_x - V²_0)/2a

a = (V²_x - V²_1)/2S_1 = (0 - 20²)/2*40 = -400/80 = -5 m/s²

S_2 = (0 - 25²)/2*(-5) = -625/-10 = 62,5 m

Answer 2

Ответ:

62.5

Объяснение:

V₁ = 20 m/s

S₁ = 40m

V₂ = 25 m/s

V = 0

S₂ = ?

S = (V²-V²₀)/2a

S₁ = -V₁²/2a

S₂ = -V₂²/2a

S₂/S₁ = (V₂/V₁)²

S₂ = S₁* (V₂/V₁)² = 40 m*(25 m/s / 20 m/s)² = 62.5 m

Answer 3

If the car undergoes the same acceleration in both cases, we can use the following equation to relate the braking distance to the initial velocity and acceleration:

$S = \frac{v^2}{2a}$

Where:

• $S$ is the braking distance
• $v$ is the velocity
• $a$ is the acceleration

Given that the braking distance $S_1$ is 40 m and the initial velocity $v_1$ is 20 m/s, we can calculate the acceleration $a$ using the equation above:

$40 = \frac{20^2}{2a}$

Solving for $a$: $a = \frac{20^2}{2 \cdot 40} = 5 \, \text{m/s}^2$

Now, we can use the calculated acceleration to find the braking distance $S_2$ at a velocity $v_2$ of 25 m/s:

$S_2 = \frac{v_2^2}{2a} = \frac{25^2}{2 \cdot 5} = 31.25 \, \text{m}$

So, the braking distance $S_2$ at a velocity $v_2$ of 25 m/s is 31.25 meters. However, none of the options you provided match this value exactly. It's possible there might be a calculation error or a typo in the options. The closest option to the calculated value is 31.5 meters, which is not among the provided options.