Answer:
e. 1000m
Step-by-step explanation:
Because we're given the initial velocity of the car (10 km/h), the final velocity (0 km/h), and the displacement between the change of these two velocities, we can use the following kinematic formula:
[tex]v_f^2 = v_0^2 + 2a\Delta x[/tex]
Where:
In this specific problem, v(f) and a are constants. Because of this, we can construct a function where the independent variable is v(0) and the dependent variable is Delta x. After all, we're trying to find how the change in the initial velocity affects the displacement of a vehicle that always comes to a stop over a constant deceleration rate.
To do this, we'll solve the given formula for the displacement.
[tex]v_f^2 = v_0^2 + 2a\Delta x \text{ //}-v_0^2\\v_f^2 - v_0^2 = 2a\Delta x \text{ //}\div 2a\\\\\to \Delta x = \frac{v_f^2 - v_0^2}{2a}[/tex]
Let a be the constant deceleration of the car. We'll substitute v(f) = 0 and a = a into the function:
[tex]\Delta x = \frac{-v_0^2}{2a}[/tex]
We want to examine how some constant change, k, to the initial velocity affects the final displacement of the car. To do this, we'll substitute v0 = k * v0.
[tex]\Delta x = \frac{-(v_0 \times k)^2}{2a}\\\\\Delta x = \frac{-v_0^2 \times k^2}{2a} = k^2 \times \frac{-v_0^2}{2a} = k^2 \times \Delta x_\text{before incrementing times k}[/tex]
From this substitution, we can see that by incrementing the initial velocity times k, we've incremented the displacement times k^2.
Thus, if we increment the speed of the car from 10 km/h to 100 km/h, then k = 10 (as 100/10 = 10).
[tex]\Delta x_{\text{100 km/h}} = \Delta x_\text{10 km/h} \times k^2 = 10 \times 10^2 = 10^3 = 1000 \text{ meters}[/tex]
The displacement of the car is 1000 meters.
Answer:
e) 1000m
Step-by-step explanation:
A car travels at a speed of 10 km/h, and upon seeing a road sign it brakes and comes to a stop after 10 metres.
To find the acceleration of the car, we can use the Constant Acceleration Equations (SUVAT).
In this case:
To find the acceleration (a) in km/h², we can substitute the values of s, u and v into the SUVAT equation v² = u² + 2as, and solve for a:
[tex]v^2=u^2+2as\\\\\\0^2=10^2+2a(0.01)\\\\\\0 = 100 + 0.02a\\\\\\-0.02a = 100\\\\\\\dfrac{-0.02a}{-0.02}=\dfrac{100}{-0.02}\\\\\\a = -5000\; \rm km/h^2[/tex]
Therefore, the acceleration of the car was -5000 km/h². (As this is negative, it corresponds to a deceleration of 5000 km/h²).
If the car had initially been travelling at 100 km/h and it had decelerated at the same rate, then:
To find displacement (s) in kilometres, substitute the values of u, v and a into the SUVAT equation v² = u² + 2as, and solve for s:
[tex]v^2 = u^2 + 2as\\\\\\0^2=100^2+2(-5000)s\\\\\\0 = 10000 -10000s\\\\\\10000s = 10000\\\\\\s = 1\; \text{km}[/tex]
Finally, convert 1 km to metres by multiplying it by 1000:
[tex]\rm1\;km= 1000\; m[/tex]
Therefore, if the car had initially been travelling at 100 km/h and it had decelerated at the same rate, it would have taken 1000 m to come to a stop.
