Question

A school bus heading east through a small town accelerates as it passes the sign post at s = 0. Marking the city limits. Its acceleration is constant 5 m/s^2 at time 0; it is 5 m east of the (+) sign post and has a velocity of 3 m/s (a) Find his position and velocity at 2 seconds. (B) where is it when its velocity is 5 m/s.

Answer 1

Answer on Question#53488 - Physics - Mechanics - Kinematics - Dynamics

A school bus heading east through a small town accelerates as it passes the sign post at $s = 0$. Marking the city limits. Its acceleration is constant $a = 5\frac{\mathrm{m}}{\mathrm{s}^2}$ at time 0; it is $l_0 = 5\mathrm{m}$ east of the (+) sign post and has a velocity of $v_{f} = 3\frac{\mathrm{m}}{\mathrm{s}}$ (a) Find his position and velocity at 2 seconds. (B) where is it when its velocity is $v_{B} = 5\frac{\mathrm{m}}{\mathrm{s}}$.

Solution:

(a) To find bus's velocity at $t = 2$ s we'll use the following formula

$$v _ {f} - v _ {i} = a t,$$

where $v_{i}$ - initial velocity, $v_{f}$ - final velocity. Since at $t = 0$ s the velocity is $v_{i} = 3\frac{\mathrm{m}}{\mathrm{s}}$, then

$$v _ {f} = v _ {i} + a t = 3 \frac {\mathrm {m}}{\mathrm {s}} + 5 \frac {\mathrm {m}}{\mathrm {s} ^ {2}} \cdot 2 \mathrm {s} = 1 3 \frac {\mathrm {m}}{\mathrm {s}}$$

Its position is defined by

$$s (t) = l _ {0} + v _ {i} \cdot t + \frac {a t ^ {2}}{2}$$

Therefore

$$s (2) = 5 \mathrm {m} + 3 \frac {\mathrm {m}}{\mathrm {s}} \cdot 2 \mathrm {s} + \frac {5 \frac {\mathrm {m}}{\mathrm {s} ^ {2}} \cdot (2 \mathrm {s}) ^ {2}}{2} = 2 1 \mathrm {m}$$

(b) Its displacement $\Delta s$ (as it accelerates from $v_{i}$ to $v_{B}$) can be found from the following formula

$$v _ {B} ^ {2} - v _ {i} ^ {2} = 2 a \Delta s$$

Thus

$$\Delta s = \frac {v _ {B} ^ {2} - v _ {i} ^ {2}}{2 a} = \frac {\left(5 \frac {\mathrm {m}}{\mathrm {s}}\right) ^ {2} - \left(3 \frac {\mathrm {m}}{\mathrm {s}}\right) ^ {2}}{2 \cdot 5 \frac {\mathrm {m}}{\mathrm {s} ^ {2}}} = 1. 6 \mathrm {m}$$

The final position is then given by

$$s = l _ {0} + \Delta s = 5 \mathrm {m} + 1. 6 \mathrm {m} = 6. 6 \mathrm {m}$$

Answer:

(a) $v_{f} = 13\frac{\mathrm{m}}{\mathrm{s}}$

$$s (2) = 2 1 \mathrm {m}$$

(b) $s = 6.6\mathrm{m}$

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