Question

An object accelerates uniformly from rest to a speed of
50 meters per second in 5.0 seconds. The average speed of
the object during the 5.0-second interval is
A 5.0 m/s
B 10 m/s
C 25 m/s
D 50 m/s

Answer 1

Answer:

25

Explanation:

v=u+at.

50=a(5)

a=10m/s^2.

distance= ut+ 1/2 at^2

=125

average speed = total distance/total time

=125/5

=25

Answer 2

The average speed of  the object during the 5.0-second interval as it accelerates uniformly from rest is 25m/s.

Hence, Option C) 25 m/s is the correct answer.

Given the data in the question;

Since the object was initially at rest,

• Initial velocity; $u = 0$

• Final velocity; $v = 50m/s$

• Time; $t = 5.0s$

First we determine the acceleration "a", using the First equation of motion:

$v = u + at$

Where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

We substitute our values into the equation

$50m/s = 0 + ( a \ * \ 5.0s)\\\\a = \frac{50m/s}{5.0s} \\\\a = 10m/s^2$

Next we Calculate the distance travelled by the object

From the Second Equation of Motion:

$s = ut \ + \ \frac{1}{2}at^2$

Where, u is the initial velocity, a is the acceleration and t is the time and s is the distance travelled.

Wes substitute our values into the equation

$s = (0\ * 5.0s) + ( \frac{1}{2} \ *\ 10m/s^2 \ * (5.0s)^2)\\\\s = \frac{1}{2} \ *\ 10m/s^2 \ * 25s^2\\\\s = 125m$

Now, we can get the Average speed of the object.

We know that, Speed is the distance travelled per unit time.

$Speed = \frac{Distance}{Time}$

We substitute our values into the equation

$Speed = \frac{125m}{5.0s}\\\\Speed = 25m/s$

The average speed of  the object during the 5.0-second interval as it accelerates uniformly from rest is 25m/s.

Hence, Option C) 25 m/s is the correct answer.

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