2. (9 points) A car starts from 10 mph and accelerates along a level road, i.e., no grade change. At 500 ft from its starting po

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2. (9 points) A car starts from 10 mph and accelerates along a level road, i.e., no grade change. At 500 ft from its starting po

int, a radar gun measures its speed as 50 mph. Assuming the car had a constant rate of acceleration, (a) calculate the time elapsed between when the car started at 10 mph to when its speed was measured and (b) what will the speed of the car be another 500 ft downstream of this point

Physics

1 answer:

Sonja [21] · Answer 1 · 2022-08-10T18:48:16+03:00

Answer:

a) t = 11.2 s

b) v = 70.5 mph

Explanation:

a)

Since we need to find the time, we could use the definition of acceleration (rearranging terms) as follows:

$t = \frac{v_{f} - v_{o}}{a} (1)$

where vf = 50 mph, and v₀ = 10 mph.
However, we still lack the value of a.
Assuming that the acceleration is constant, we can use the following kinematic equation:

$v_{f} ^{2} - v_{o} ^{2} = 2*a* \Delta x (2)$

Since we know that Δx = 500 ft, we could solve (2) for a.
In order to simplify things, let's first to convert v₀ and vf from mph to m/s, as follows:

$v_{o} = 10 mph*\frac{1609m}{1mi} *\frac{1h}{3600s} = 4.5 m/s (3)$

$v_{f} = 50 mph*\frac{1609m}{1mi} *\frac{1h}{3600s} = 22.5 m/s (4)$

We can do the same process with Δx, from ft to m, as follows:

$\Delta x = 500 ft *\frac{0.3048m}{1ft} = 152.4 m (5)$

Replacing (3), (4), and (5) in (2) and solving for a, we get:

$a = \frac{v_{f} ^{2} - v_{o}^{2}}{2*\Delta x} = \frac{(22.5m/s) ^{2} - (4.5m/s)^{2}}{2*152.4m} = 1.6 m/s2 (6)$

Replacing (6) in (1) we finally get the value of the time t:

$t = \frac{v_{f} - v_{o}}{a} = \frac{(22.5m/s) - (4.5m/s)}{1,6m/s2} = 11.2 s (7)$

b)

Since the acceleration is constant, as we know the displacement is another 500 ft (152.4m), if we replace in (2) v₀ by the vf we got in a), we can find the new value of vf, as follows:

$v_{f} = \sqrt{v_{o} ^{2} +( 2*a* \Delta x)} = \sqrt{(22.5m/s)^{2} + (2*1.6m/s2*152.4m)} \\ v_{f} = 31.5 m/s (8)$