All categories

2 years ago

A truck traveling with an initial velocity of 22m/s comes to a stop in 17.32 secs. What is the acceleration of the truck?

Physics

2 answers:

Tomtit [17]2 years ago

6 0

Explanation:

Given:

v₀ = 22 m/s

v = 0 m/s

t = 17.32 s

Find: a

v = at + v₀

(0 m/s) = a (17.32 s) + (22 m/s)

a = -1.270 m/s²

Round as needed.

trasher [3.6K]2 years ago

6 0

<u>Answer:</u>

A truck traveling with an initial velocity of 22 m/s comes to a stop in 17.32 sec, the acceleration of the truck is $- 1.27 m/s^2$

<u>Explanation</u>

From the Given statements, we know that Initial Velocity (u) = 22 m/s

time (t) = 17.32 seconds

Final Velocity (v) = 0 m/s [ because the truck Stops]

find acceleration on Applying the Equation of Motion

v = u + at

substituting the values in above equation, we find

a = $\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}$

= $\frac{0-22}{17.32} \mathrm{m} / \mathrm{s}^{2}$

a = $- 1.27 m/s^2$ [negative sign shows reverse direction]

You might be interested in

A car moves with a speed 72km/h, the driver uses the brakes, the car stops after 8 seconds, calculate its speed after 10 seconds

vampirchik [111]

The acceleration of the car is solved by subtracting the initial speed from the final speed then dividing the result by the elapsed time.

initial speed = 72 km/hr = 20 m/s

final speed = 0 m/s

elapsed time = 5 seconds

acceleration = (0 m/s – 20 m/s) / 5 s

acceleration = - 20m/s / 5 s

acceleration = -4 m/s^2

8 0

3 years ago

7. Which best describes the energy change th pushes a large rock down a hill?

svlad2 [7]

Answer:

Explanation:

Gravitational Energy is the energy of position or place. A rock resting at the top of a hill contains gravitational Potential energy. Hydropower, such as water in a reservoir behind a dam, is an example of gravitational potential energy.

6 0

2 years ago

Read 2 more answers

Which is the correct answer ?

hodyreva [135]

D ............................

7 0

3 years ago

Read 2 more answers

Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has a magnitude of 3.7m/s2. We want

lbvjy [14]

Explanation:

It is given that, Onur drops a basketball from a height of 10 m on Mars, where the acceleration due to gravity has a magnitude of 3.7 m/s².

The second equation of kinematics gives the relationship between the height reached and time taken by it.

Here, the ball is droped under the action of gravity. The value of acceleration due to gravity on Mars is positive.

We want to know how many seconds the basketball is in the air before it hits the ground. So, the formula is :

$\Delta x=v_ot+\dfrac{1}{2}at^2$

t is time taken by the ball to hit the ground

$v_o$ is initial speed of the ball

So, the correct option is (A).

8 0

3 years ago

A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t

Charra [1.4K]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time

V = at + U

using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane

$V_{x} = mt + V_{o}$ where m is the slope

Comparing equation (1) and (2)

$V = V_{x}$

a = m

$U = V_{o}$

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

a = - 1.40 m/s²

Sunstituting a = - 1.40 m/s² and u = 22 m/s

$V_{x} = -1.40t + 22$