Question

an object starts from rest with a constant acceleration of 8.0m/s^2 along a straight line.find (a)the speed at the end of 5.0s (b)the average speed for 5s interval and (c)the distance travelled in the 5.0s

Answer 1

Initially object is in rest, u = 0 m/s
Acceleration (a) = v-u/t = 8 m/s^2

(a)

Time is (t) = 5 sec

Acceleration (a) = v-u/t 
          8 m/s^2 = v-0/5 sec
                  v = 40 m/s

(b)

Average speed = (0+40m/s) /5 = 8 m/s

(c)

Distance traveled (D) = s x t
                                    = 40 m/s x 5
                                    = 200 m

Answer 2

Given:

Acceleration $=8m/s^{2}$

To find:

1. The speed at the end of $5s.$
2. The average speed during the $5s$ interval.
3. The distance traveled in $5s$.

Solution:

Step 1

We have been given that the ball starts from rest with an acceleration of $8m/s^{2}$ . Hence, the final velocity of the object at the end of 5 s will be given by

$v=u+at$

We have,

$u=0m/s$     $;$ $a=8m/s^{2}$     $;$  $t=5s$

We get

$v=(0)+8(5)$

$v=40m/s$

Hence, the final velocity of the object after 5 s will be 40 m/s.

Step 2

We know, average speed of the journey is the total distance traveled by an object divided by the total time of the journey.

Mathematically,   $V_{avg}=\frac{total-distance-traveled}{total-time-taken}$

We have,

$u=0m/s$    $;$ $v=40m/s$    $;$ $a=8m/s^{2}$

Hence, the distance traveled by the object in 5 s will be given by

$v^{2}- u^{2}=2as$

$(40)^{2}- (0)^{2}=2(8)s$

$s=100m$

Hence, the object travels a distance of 100 m in 5 seconds.

Now,

$V_{avg}=\frac{total-distance-traveled}{total-time-taken}$

$V_{avg}=\frac{100}{5}$

$V_{avg}=20m/s$

Therefore, the average speed of the object throughout the journey is 20 m/s.

Final answer:

Hence,

1. The speed of the object at the end of 5 s will be 40 m/s.
2. The average speed of the object throughout the journey is 20 m/s.
3. The object travels 100 m in 5 seconds.