Question

An object moves with constant acceleration 4.15 m/s2 and over a time interval reaches a final velocity of 10.4 m/s.
(a) If its original velocity is 5.20 m/s, what is its displacement during the time interval?
m

(b) What is the distance it travels during this interval?


(c) If its initial velocity is
−5.20 m/s,
what is its displacement during this interval?


(d) What is the total distance it travels during the interval in part (c)?

Answer 1

Answer on Question #45691 – Physics – Mechanics | Kinematics | Dynamics

An object moves with constant acceleration $4.15 \, \text{m/s}^2$ and over a time interval reaches a final velocity of $10.4 \, \text{m/s}$.

(a) If its original velocity is $5.20 \, \text{m/s}$, what is its displacement during the time interval?

m

(b) What is the distance it travels during this interval?

(c) If its initial velocity is

$-5.20 \, \text{m/s}$

what is its displacement during this interval?

(d) What is the total distance it travels during the interval in part (c)?

Solution:

(a)

$V_1 = 5.2 \frac{\text{m}}{\text{s}}$ – initial velocity of the object;

$V_2 = 10.4 \frac{\text{m}}{\text{s}}$ – final velocity of the object;

$a = 4.15 \frac{\text{m}}{\text{s}^2}$ – acceleration;

T – time of travelling;

d – displacement during time T;

Rate equation for the object:

$$V_2 = V_1 + aT$$ $$T = \frac{V_2 - V_1}{a}$$

Equation of motion for the object:

$$\begin{aligned} d &= V_1 T + \frac{aT^2}{2} = T \left( V_1 + \frac{aT}{2} \right) = \frac{V_2 - V_1}{a} \left( V_1 + \frac{V_2 - V_1}{2} \right) = \frac{V_2 - V_1}{a} \left( \frac{V_1 + V_2}{2} \right) = \\ &= \frac{V_2^2 - V_1^2}{2a} = \frac{ \left( 10.4 \frac{\text{m}}{\text{s}} \right)^2 - \left( 5.2 \frac{\text{m}}{\text{s}} \right)^2 }{ 2 \cdot 4.15 \frac{\text{m}}{\text{s}^2} } = 9.8 \, \text{m} \end{aligned}$$

(b)

Here the distance = displacement = 9.8 m because there wasn't change in moving direction during time T.

(c)

$V_3 = -5.2 \frac{\text{m}}{\text{s}}$ – initial velocity of the object;

$V_4 = 10.4 \frac{\text{m}}{\text{s}}$ – final velocity of the object;

$t_2$ – time of travelling;

$d_2$ – displacement during time $t_2$;

$$V_4 = V_3 + a t_2$$ $$t_2 = \frac{V_4 - V_3}{a}$$

Equation of motion for the object:

$$\begin{aligned} d = V_3 t_2 + \frac{a t_2^2}{2} &= t_2 \left(V_3 + \frac{a t_2}{2}\right) = \frac{V_4 - V_3}{a} \left(V_3 + \frac{V_4 - V_3}{2}\right) = \frac{V_4 - V_3}{a} \left(\frac{V_3 + V_4}{2}\right) = \\ &= \frac{V_4^2 - V_3^2}{2a} = \frac{\left(10.4 \frac{\mathrm{m}}{\mathrm{s}}\right)^2 - \left(-5.2 \frac{\mathrm{m}}{\mathrm{s}}\right)^2}{2 \cdot 4.15 \frac{\mathrm{m}}{\mathrm{s}^2}} = 9.8 \mathrm{~m} \end{aligned}$$

(d)

$a = 4.15 \frac{\mathrm{m}}{\mathrm{s}^2}$ – acceleration;

$V_1 = -5.2 \frac{\mathrm{m}}{\mathrm{s}}$ – initial velocity of the object;

$t_1$ – time of first part travelling;

$d_1$ – displacement during time $t_2$;

$V_2 = 0$ – zero velocity;

$t_2$ – time of second part travelling;

$V_3 = 10.4 \frac{\mathrm{m}}{\mathrm{s}}$ – final velocity of the object;

The total distance here will be different, thus we can split the movement of the object into 2 parts:

1) when it starts at $-5.2 \frac{\mathrm{m}}{\mathrm{s}}$ then stops to change direction

Rate equation of the object:

$$V_2 = V_1 + a t_1 = 0$$ $$t_1 = \frac{V_1}{a}$$

Equation of motion for the object:

$$d_1 = V_1 t_1 - \frac{a t_1^2}{2} = \frac{V_1^2}{a} - \frac{V_1^2}{2a} = \frac{V_1^2}{2a} = \frac{\left(-5.2 \frac{\mathrm{m}}{\mathrm{s}}\right)^2}{2 \cdot 4.15 \frac{\mathrm{m}}{\mathrm{s}^2}} = 3.26 \mathrm{~m}$$

2) the distance object passes till its velocity becomes $V_3 = 10.4 \frac{\mathrm{m}}{\mathrm{s}}$

Rate equation of the object:

$$V_3 = V_2 + a t_2$$ $$t_2 = \frac{V_3}{a}$$

Equation of motion for the object:

$$d _ {2} = V _ {2} t _ {2} + \frac {a t _ {2} ^ {2}}{2} = 0 + \frac {a t _ {2} ^ {2}}{2} = \frac {a}{2} \left(\frac {V _ {3}}{a}\right) ^ {2} = \frac {V _ {3} ^ {2}}{2 a} = \frac {\left(1 0 . 4 \frac {m}{s}\right) ^ {2}}{2 \cdot 4 . 1 5 \frac {m}{s ^ {2}}} = 1 3. 0 3 m$$

Total distance = 3.26 m + 13.03 m = 16.3 m

Answer: a) 9.8 m

b) 9.8 m

c) 9.8 m

d) 16.3 m

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