Question

a = 4 – 2t Initial velocity at t=0,u=5, find
distance travelled till 12sec?
(A) 294.6 m
(B) 282.6 m
(C) 292.8 m
(D) 296.6 m​

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

a = 4 – 2t Initial velocity at t=0,u=5, find distance travelled till 12sec

(A) 294.6 m

(B) 282.6 m

(C) 292.8 m

(D) 296.6 m

EVALUATION

Here it is given that

$\sf{a = 4 - 2t}$

$\displaystyle \sf{ \frac{dv}{dt} = 4 - 2t}$

$\displaystyle \sf{ \implies \: dv =( 4 - 2t)dt}$

Integrating both sides we get

$\displaystyle \sf{ \implies \: v =4t - {t}^{2} + c }$

Now at t = 0 we have v = u = 5

Thus we get c = 5

$\displaystyle \sf{ \implies \: v =4t - {t}^{2} + 5 }$

$\displaystyle \sf{ \implies \: \frac{ds}{dt} =4t - {t}^{2} + 5 }$

$\displaystyle \sf{ \implies \: ds =(4t - {t}^{2} + 5)dt }$

$\displaystyle \sf{ \implies \: s =2 {t}^{2} - \frac{ {t}^{3} }{3} + 5t + k }$

At t = 0 we have s = 0

Thus we get

$\displaystyle \sf{ \implies \: s =2 {t}^{2} - \frac{ {t}^{3} }{3} + 5t }$

For t = 12 we get

$\displaystyle \sf{ \implies \: s =2 \times {12}^{2} - \frac{ {12}^{3} }{3} +( 5 \times 12) }$

$\displaystyle \sf{ \implies \: s = - 228 }$

Again for t = 5 we have

$\displaystyle \sf{ \implies \: s =2 \times {5}^{2} - \frac{ {5}^{3} }{3} +( 5 \times 5) }$

$\displaystyle \sf{ \implies \: s = \frac{100}{3} }$

Hence the required distance

$\displaystyle \sf{ = 228 + 2 \times \frac{100}{3} \: \: \: m}$

$\displaystyle \sf{ = 294.6 \: \: \: m}$

FINAL ANSWER

Hence the correct option is (A) 294.6 m

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Answer 2

Given :  a = 4 – 2t Initial velocity at t=0,u=5,

To find :  

distance travelled till 12sec

(A) 294.6 m

(B) 282.6 m

(C) 292.8 m

(D) 296.6 m​

Solution:

a = 4 – 2t

a = dv/dt

=> dv/dt = 4 - 2t

=> dv = (4 - 2t)dt

integrating both sides

=> v = 4t  - t² +  c

at t=0,u=5

=> 5 = 0 - 0 + c

=> c = 5

Hence v  = 4t  - t² +  5

v = ds/dt

=> ds = (4t  - t² +  5) dt

v = -t² + 4t + 5

=> v = -t² -t + 5t + 5

=> v = -t(t + 1) + 5(t + 1)

=> v = (t + 1)( 5 - t)

v > 0  if     -1 < t < 5

v < 0  if   t < - 1  or  t  > 5

Hence from 0 to 12 secs

0 to 5 secs velocity is in +ve direction

and from 5 to 12 sec velocity is in -ve direction

distance covered in always + ve

so need to split the distance covered in interval from 0 to 5 and 5 to 12.

Hence  

ds = (4t  - t² +  5) dt

Integrating both sides

$s=| \int\limits^5_0 { (4t - t^2 + 5} \, dt |+ | \int\limits^{12}_5 { (4t - t^2 + 5} \, dt |$

$s=| \left 2t^2 - \frac{t^3}{3} + 5t \right|_0^5 |+ | \left 2t^2 - \frac{t^3}{3} + 5t \right|_5^{12} |$

s =  | (50 - 125/3 + 25 - 0) | + | 288 - 576 + 60 - (50 - 125/3 + 25 - 0) |

=> s = |  100/3 |  + |  -228 - 100/3|

=>s = 100/3 + 228 + 100/3

=> s = 228 + 200/3

=> s = 228 + 66.67

=> s = 294.67

distance travelled till 12sec = 294.67 m

from given option 294.6 m is closest.

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