a truck, initially traveling at a speed of 22 meters per second, increases speed at a constant rate of 2.4 meters per second2 fo

Question

mafiozo [28]

4 years ago

13

a truck, initially traveling at a speed of 22 meters per second, increases speed at a constant rate of 2.4 meters per second2 fo

r 3.2 seconds. what is the total distance traveled by the truck during this 3.2-second time interval?

Physics

2 answers:

sineoko [7] · Answer 1 · 2021-12-12T14:03:40+03:00

sineoko [7]4 years ago

6 0

U = 22 m/s, a = 2.4 m/s², t = 3.2 s

s = ut + (1/2)at²

s = 22*3.2 + (1/2)*2.4*3.2²

s = 70.4 + 12.228

s = 82.688 m.

Len [333] · Answer 2 · 2021-12-12T15:18:08+03:00

The total distance travel by the truck during the time interval of and with the uniform acceleration is $\boxed{82.688{\text{ m}}}$ or $\boxed{82.69{\text{ m}}}$ or $\boxed{82.7{\text{ m}}}$ .

Further explain:

We have to calculate the distance travel during time interval of $3.2{\text{ s}}$ .

Given:

Initial velocity of the truck is $22{{{\text{ m}}} \mathord{\left/ {\vphantom {{{\text{ m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$

The constant rate of increase of velocity, which is also known as uniform acceleration of the truck is $2.4{{{\text{ m}}} \mathord{\left/ {\vphantom {{{\text{ m}}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}}$ .

Time duration of the motion of the truck is $3.2{\text{ s}}$ .

Formula and concept used:

For calculating the distance traveled by the truck during the time interval of $3.2{\text{ s}}$ .

Use the kinematic equation of the motion with uniform acceleration as,

$S = ut + \dfrac{1}{2}a{t^2}$ …… (1)

Here, $S$ is the distance travelled by the truck.

$u$ is the initial velocity of the truck.

$a$ is the acceleration of the truck or constant rate of change of acceleration.

$t$ is the time duration of the motion of the truck with uniform acceleration.

Calculation:

Substitute the value of $u$ as $22{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ , value of $a$ as $3.2{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}}$ and the value of $t$ as $3.2{\text{ s}}$ in the equation (1).

$\begin{gathered}S=22 \times 3.2 + \frac{1}{2} \times 2.4 \times {3.2^2} \hfill\\S=70.4 + 12.288\hfill\\S=82.688{\text{ m}} \hfill\\S\approx 82.7{\text{ m}}\hfill\\\end{gathered}$

So, the truck will travel the distance of $82.7{\text{ m}}$ during the time interval of $3.2{\text{ s}}$ with the uniform acceleration of $2.4{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}}$ .