Physics, asked by tbs01, 7 months ago

If x=t^3 - 4t^2 - 3t then find the distance traveled from t=0 to t=4s.
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Answers

Answered by nirman95
54

Given:

x=t^3 - 4t^2 - 3t

To find:

The distance traveled from t=0 to t=4s.

Calculation:

Since the question asks for "distance" and not "displacement" , we need to first check that if the body has stopped and changed its direction of motion in the given time interval.

x =  {t}^{3}  - 4 {t}^{2}  - 3t

 \implies \: v =  \dfrac{dx}{dt}

 \implies \: v =  \dfrac{d( {t}^{3}  - 4 {t}^{2}  - 3t)}{dt}

 \implies \: v = 3 {t}^{2}  - 8t - 3

Now, put v = 0 , to check if the car stopped.

 \implies \: v = 3 {t}^{2}  - 8t - 3 = 0

 \implies \:  3 {t}^{2}  - 8t - 3 = 0

 \implies \:  3 {t}^{2}  - (9 - 1)t - 3 = 0

 \implies \:  3 {t}^{2}  - 9t + t- 3 = 0

 \implies \:  3t(t - 3)  + 1( t- 3 )= 0

 \implies \:  (3t + 1)( t- 3 )= 0

So, t = 3 sec , at which the car stopped.

In that case, first we need to calculate the position at t = 3 sec and t = 0 sec and t =4 sec.

 \therefore \:  x \bigg|_{t = 3 \: sec} =  {3}^{3}  - 4 ({3}^{2} ) - 3(3)

 \implies \:  x \bigg|_{t = 3 \: sec} =  27  - 36- 9

 \implies \:  x \bigg|_{t = 3 \: sec} =  - 18 \: m

At t = 0 sec ,

 \therefore \:  x \bigg|_{t = 0\: sec} =  {0}^{3}  - 4 ({0}^{2} ) - 3(0)

 \implies\:  x \bigg|_{t = 0\: sec} = 0 \: m

At t = 4 sec,

 \therefore \:  x \bigg|_{t = 4\: sec} =  {4}^{3}  - 4 ({4}^{2} ) - 3(4)

 \implies\:  x \bigg|_{t = 4\: sec} =  64- 64 - 12

 \implies\:  x \bigg|_{t = 4\: sec} =  - 12 \: m

So, the particle move in the sequence as follows:

  • At t = 0 , it was at 0 m position.

  • At t = 3 sec , it was at -18 m position, and it stopped and changed its direction (travelled 18 m)

  • At t = 4 sec , it is at -12 m position. (travelled 6 metres from -18 m position).

So, net distance will be :

 \therefore d = 18 + 6 = 24 \: m

So, particle travelled for 24 metres in the given time interval.

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