A jogger accelerates from rest to 2.78 m/s in 2.79 s. A car accelerates from 21.8 to 37.5 m/s also in 2.79 s. (a) Find the magnitude of the acceleration of the jogger. (b) Determine the magnitude of the acceleration of the car. (c) How much further does the car travel than the jogger during the 2.79 s?
To find the magnitude of acceleration, we can use the equation:
Acceleration (a) = Change in velocity (Δv) / Time taken (Δt)
For the jogger:
(a) Δv (jogger) = Final velocity (vjogger) - Initial velocity (0) = 2.78 m/s - 0 m/s = 2.78 m/s
Δt (jogger) = 2.79 s
Substituting these values into the equation, we get:
(a) a (jogger) = Δv (jogger) / Δt (jogger)
= 2.78 m/s / 2.79 s
≈ 0.996 m/s²
Therefore, the magnitude of the jogger's acceleration is approximately 0.996 m/s².
For the car:
(a) Δv (car) = Final velocity (vcar) - Initial velocity (21.8 m/s) = 37.5 m/s - 21.8 m/s = 15.7 m/s
Δt (car) = 2.79 s
Substituting these values into the equation, we get:
(a) a (car) = Δv (car) / Δt (car)
= 15.7 m/s / 2.79 s
≈ 5.629 m/s²
Therefore, the magnitude of the car's acceleration is approximately 5.629 m/s².
To find the distance traveled by each:
For the jogger:
The distance traveled (djogger) = Initial velocity (0) * Time (Δt) + 0.5 * Acceleration (a) * Time² (Δt²)
= 0 * 2.79 s + 0.5 * 0.996 m/s² * (2.79 s)²
= 0 + 0.5 * 0.996 m/s² * 7.7841 s²
= 0 + 3.872 m
= 3.872 m
For the car:
The distance traveled (dcar) = Initial velocity (21.8 m/s) * Time (Δt) + 0.5 * Acceleration (a) * Time² (Δt²)
= 21.8 m/s * 2.79 s + 0.5 * 5.629 m/s² * (2.79 s)²
= 60.702 m + 0.5 * 5.629 m/s² * 7.7841 s²
= 60.702 m + 0.5 * 5.629 m/s² * 60.58939921 s²
= 60.702 m + 0.5 * 338.488596 m²/s²
= 60.702 m + 169.244 m²/s²
= 229.946 m
Therefore, the car travels approximately 229.946 m and the jogger travels approximately 3.872 m during the 2.79 s.