Question

A body at rest is given an initial uniform acceleration of 8.0ms for 30s after which the acceleration is reduced to 5.0ms for the next 20s.the body maintains the speed attained for 60s after which is brought to rest in 20s.draw the velocity time graph using the information above.calculate (a).maximum speed attained during the motion. (B).average retardation as the body is being brought to rest. (C).total distance travelled during the first 50s. (D).average spped during the same interval as in (c).

Answer 1

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Answer 2

Answer:

(A). The maximum speed attained during the motion is 340 m/s.

(B).The average retardation as the body is being brought to rest is $17 m/s^{2}$.

(C).The total distance traveled in first 50 seconds is  9400 m.

(D).Average speed during the same interval as in (c) is 188 m/s.

Explanation:

Given:

Initial uniform acceleration of $8.0 \ m/s^{2}$ for 30 s after which the acceleration is reduced to $5.0 \ m/s^{2}$ for the next 20 s.

The body maintains the speed attained for 60 s after which is brought to rest in 20 s.

To Find:

(a).maximum speed attained during the motion. (B).average retardation as the body is being brought to rest. (C).total distance traveled during the first 50s. (D).average speed during the same interval as in (c).

Formula Used:

The equation for law of motion.

$v=u+at$   ------------- formula no.01.

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

Where, v is the final velocity, u is the initial velocity,s is distance  and t is the time.

Solution:

Apply  formula no.01  for t=0 to 30 sec at acceleration  $8.0 \ m/s^{2}$.

$v=0+8\times 30\\ = 240 m/s^{2}$

Apply  formula no.01  for t=30 to 50 sec at acceleration  $5.0 \ m/s^{2}$.

$v=8\times 30+5\times 20\\ = 240+100= 340 m/s^{2}$

The body remains the same velocity for next 60 seconds.

Therefore, $v=340 \ m/s^{2}$.

When the body is brought to rest at $t=130 \ seconds$. the $v=0.$

The attached the velocity time graph by  using  above calculated information.

(a).Maximum speed attained during the motion.

As per attached  velocity time graph graph, the maximum speed attained is 340 meters per second.

Thus, the maximum speed attained during the motion is 340 m/s.

(B).Average retardation as the body is being brought to rest.

    Average retardation of the body when it is brought to rest is as per following.

$0=340+a\times 20\\a=-\frac{340}{20} = -17 m/s^{2}$

  Thus, the average retardation as the body is being brought to rest is $17 m/s^{2}$.

(C).Total distance traveled during the first 50 s.  

     The distance traveled in first 30 seconds, $s_{1}$.

      Applying formula no.02.

       $s_{1} =ut+\frac{1}{2} at^{2}$

             $=0\times t+\frac{1}{2} \times 8\times 30^{2}\\= 3600\ m$

     The distance traveled in 30 to 50 seconds,$s_{2}$.

      Applying formula no.02.

        $s_{2} =240\times20+\frac{1}{2}\times 5\times 20^{2}$

             $=5800\ meter.$

   The total distance traveled in 50 seconds, $s=s_{1} +s_{2}$

                                                                              $=3600+5800=9400\ m.$

   Thus,the total distance traveled in first 50 seconds is  9400 m.

(D). Average speed during the same interval as in (c).

         Average speed for first 50 seconds $=\frac{ Distance\ traveled\ in\ 50 \ seconds}{Time\ taken}$

                                                                       $=\frac{9400}{50} = 188\ m/s$

     Thus,average speed during the same interval as in (c) is 188 m/s.

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