Question

An automobile driver increasing the speed at a constant rate from 45 km/h to 58 km/h in 0.49 min. A bicycle rider speeds up at a constant rate from rest to 13 km/h in 0.49 min. What are the magnitudes of (a) the driver's acceleration and (b) the rider's acceleration?

Answer 1

Answer on Question #51384, Physics, Mechanics | Kinematics | Dynamics

Question

An automobile driver increasing the speed at a constant rate from 45 km/h to 58 km/h in 0.49 min. A bicycle rider speeds up at a constant rate from rest to 13 km/h in 0.49 min. What are the magnitudes of (a) the driver's acceleration and (b) the rider's acceleration?

Solution

(a) First of all, will rewrite the value:

$$V _ {0} = 4 5 k m / h = \frac {4 5 0 0 0 m}{3 6 0 0 s} = 1 2. 5 m / s$$ $$V _ {1} = 5 8 k m / h = \frac {5 8 0 0 0 m}{3 6 0 0 s} = 1 6. 1 1 m / s$$ $$t = 0. 4 9 \mathrm {m i n} = 2 9. 4 \mathrm {s}$$

Since the driver increasing the speed at a constant rate:

$$V _ {1} = V _ {0} + a t \Rightarrow$$ $$a = \frac {V _ {1} - V _ {0}}{t} = \frac {1 6 . 1 1 m / s - 1 2 . 5 m / s}{2 9 . 4 s} = 0. 1 2 2 8 \frac {m}{s ^ {2}}$$

(b) Since the bicycle rider speeds up at a constant rate from rest:

$$V _ {1} = 0 + a t = a t$$ $$V _ {1} = 1 3 k m / h = 3. 6 1 m / s$$ $$\Rightarrow a = \frac {V _ {1}}{t} = \frac {3 . 6 1 m / s}{2 9 . 4 s} = 0. 1 2 2 8 \frac {m}{s ^ {2}}$$

Answer

(a) $0.1228\frac{m}{s^2}$

(b) $0.1228\frac{m}{s^2}$

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