A particle moves along a straight line- The acceleration of the particle as function of time is given by - - 6- - 12- Initial velocity of the particle is 9 - and the distance travelled by the particle in 5 - is 7-in - Find the value of -

Given, a=6t 12 u=9 m/s Distance travelled in 5 s=7x As we know. a=dvdt...(i) adt=dv Integrating both sides by putting the values we get, ∫_9^v=dv=∫^t_0(6t 12)dt ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

SHM expression

A particle mo...

Question

A particle moves along a straight line. The acceleration of the particle as function of time is given by $a = 6 t - 12$ . Initial velocity of the particle is $9 m / s$ and the distance travelled by the particle in $5 s$ is $7 x$ (in $m$ ). Find the value of $x$ .

Open in App

Solution

Given,
$a = 6 t - 12 u = 9 m / s$
Distance travelled in $5 s = 7 x$
As we know.
$a = \frac{d v}{d t} . . . (i) a d t = d v$
Integrating both sides by putting the values we get,
$\int_{9}^{v} = d v = \int_{0}^{t} (6 t - 12) d t v - 9 = 3 t^{2} - 12 t v = 3 t^{2} - 12 t + 9 = 0$
at $t = 1 s$ and $t = 3 s$

As we know,
$v = \frac{d s}{d t} . . . (i i) \frac{d s}{d t} = \int_{3}^{5} 3 t^{2} - 12 t + 9 \int_{0}^{s} d s = \int_{3}^{5} (3 t^{2} - 12 t + 9) d t$

Hence,
$s = ∣ ∣ \int (3 t^{2} - 12 t + 9)_{0}^{1} d t ∣ ∣ + ∣ ∣ \int (3 t^{2} - 12 t + 9)_{1}^{3} d t ∣ ∣ + ∣ ∣ \int (3 t^{2} - 12 t + 9)_{3}^{5} ∣ ∣ d t s = 28 m$

Therefore,
$7 x = 28 x = 4 m$

Suggest Corrections

Similar questions

Q. A particle moves along a straight line in such a way that its acceleration is increasing at the rate of

2 m / s^{3}

. Its initial acceleration and velocity were zero. The distance which it covers in the

3^{r d}

second is

x

meter. Find

3 x

Q. The velocity of a particle moving on the

x

-axis is given by

v = x^{2} + x

, where

v

is in

m / s

and

x

is in

m

. Find its acceleration (in

m / s^{2}

) when passing through the point

x = 2 m

Q. Displacement

X

of a particle at time

t

moving along a straight-line path is given by

X^{2} = a t^{2} + 2 b t + c

, where

a, b

and

c

are constants. If acceleration of the particle varies as

\frac{a - v^{n}}{X^{m}}

, where

v

is velocity of particle at time 𝑡, then find

(n + m)

Q. The displacement of a particle moving along 𝑥-axis is given by

x = a + b t + c t^{2}

. The acceleration of the particle is

Q. A particle moving along a straight line with constant acceleration is having initial and final velocity as

5 m / s

and

15 m / s

respectively in a time interval of

5 s

. Find the distance travelled by the particle and the acceleration of the particle. If the particle continues with same acceleration, find the distance covered by the particle in the

8 t h

second of its motion (direction of motion remains same).

Join BYJU'S Learning Program

Explore more

SHM expression

Standard XII Physics

Join BYJU'S Learning Program

A particle moves along a straight line. The acceleration of the particle as function of time is given by a=6t–12. Initial velocity of the particle is 9m/s and the distance travelled by the particle in 5s is 7x(in m). Find the value of x.

A particle moves along a straight line. The acceleration of the particle as function of time is given by $a = 6 t - 12$ . Initial velocity of the particle is $9 m / s$ and the distance travelled by the particle in $5 s$ is $7 x$ (in $m$ ). Find the value of $x$ .