A particle moving with an initial velocity 2.5 m/s along the positive

1.	A particle moving with an initial velocity 2.5 m/s along the positive x direction accelerates uniformly at the rate 0.50 m/s2. (i) find the distance travelled in the first 2 sec. (ii) calculate the time taken to reach the velocity of 7.5 m/s? (iii) calculate the distance travelled in reaching the velocity 7.5 m/s?
Answer» ANSWER: ${\bold{(i)Distance=6\:m}}$ ${\bold{(ii)<klux>TIME</klux>=10\:sec}}$ ${\bold{(iii)Distance=50\:m}}$ Explanation: ${\bold{\huge{\underline{SOLUTION-}}}}$ • In the given question information given about a particle moving with initial speed and uniform acceleration. • At first we have to find the distance travelled in first 2 sec. • Second, we have to find the time TAKEN to REACH velocity 7.5 m/s. • At last, we have to find the distance travelled in reaching the velocity of 7.5 m/s. $\underline \bold{Given :} \\ \\ \implies Initial \: Velocity(u) = 2.5 \: m/s \\ \\ \implies Acceleration(a) = 0.50 \: m/ {s}^{2} \\ \\ \implies Time(t) = 2 \: sec \\ \\ \underline \bold{To \: find : } \\ \\ \implies Distance \: travelled \: in \: 2 \: sec =? \\ \\ \implies Time \: taken \: to \: reach \: velocity \: 7.5 \: m/s =? \\ \\ \implies Distance \: travelled \: in \: reaching \: velocity \: 7.5 \: m/s = ?$ • According to given question : $\bold{For\:first\:question:}\\\\ \bold{By \: Third \: equation \: of \: motion : } \\ \\ \implies s = ut + \frac{1}{2}a {t}^{2} \\ \\ \implies s = 2.5 \times 2 + \frac{1}{2} \times 0.50 \times {2}^{2} \\ \\ \implies s = 5 + 0.25 \times 4 \\ \\ \implies s = 5 + 1 \\ \\ \bold {\implies s = 6 \: m}$ $\bold{For \: second \: question : } \\ \\ \bold{By \: First \: equation \: of \: motion : } \\ \\ \implies v = u + at \\ \\ \implies 7.5 = 2.5 + 0.5 \times t \\ \\ \implies 7.5 - 2.5 = 0.5t \\ \\ \implies t = \frac{5}{0.5} \\ \\ \bold { \implies t = 10 \: sec}$ $\bold{For \: third \: question : } \\ \\ \bold{By \: Second \: equation \: of \: motion : } \\ \\ \implies {v}^{2} = {u}^{2} + 2as \\ \\ \implies ({7.5})^{2} = ({2.5})^{2} + 2 \times 0.5 \times s \\ \\ \implies 56.25 = 6.25 + s \\ \\ \implies s = 56.25 - 6.25 \\ \\ \bold{\implies s = 50 \: m}$

A particle moving with an initial velocity 2.5 m/s along the positive x direction accelerates uniformly at the rate 0.50 m/s2. (i) find the distance travelled in the first 2 sec. (ii) calculate the time taken to reach the velocity of 7.5 m/s? (iii) calculate the distance travelled in reaching the velocity 7.5 m/s?

Answer»

${\bold{(i)Distance=6\:m}}$

${\bold{(ii)<klux>TIME</klux>=10\:sec}}$

${\bold{(iii)Distance=50\:m}}$

Explanation:

${\bold{\huge{\underline{SOLUTION-}}}}$

• In the given question information given about a particle moving with initial speed and uniform acceleration.

• At first we have to find the distance travelled in first 2 sec.

• Second, we have to find the time TAKEN to REACH velocity 7.5 m/s.

• At last, we have to find the distance travelled in reaching the velocity of 7.5 m/s.

$\underline \bold{Given :} \\ \\ \implies Initial \: Velocity(u) = 2.5 \: m/s \\ \\ \implies Acceleration(a) = 0.50 \: m/ {s}^{2} \\ \\ \implies Time(t) = 2 \: sec \\ \\ \underline \bold{To \: find : } \\ \\ \implies Distance \: travelled \: in \: 2 \: sec =? \\ \\ \implies Time \: taken \: to \: reach \: velocity \: 7.5 \: m/s =? \\ \\ \implies Distance \: travelled \: in \: reaching \: velocity \: 7.5 \: m/s = ?$

• According to given question :

$\bold{For\:first\:question:}\\\\ \bold{By \: Third \: equation \: of \: motion : } \\ \\ \implies s = ut + \frac{1}{2}a {t}^{2} \\ \\ \implies s = 2.5 \times 2 + \frac{1}{2} \times 0.50 \times {2}^{2} \\ \\ \implies s = 5 + 0.25 \times 4 \\ \\ \implies s = 5 + 1 \\ \\ \bold {\implies s = 6 \: m}$

$\bold{For \: second \: question : } \\ \\ \bold{By \: First \: equation \: of \: motion : } \\ \\ \implies v = u + at \\ \\ \implies 7.5 = 2.5 + 0.5 \times t \\ \\ \implies 7.5 - 2.5 = 0.5t \\ \\ \implies t = \frac{5}{0.5} \\ \\ \bold { \implies t = 10 \: sec}$

$\bold{For \: third \: question : } \\ \\ \bold{By \: Second \: equation \: of \: motion : } \\ \\ \implies {v}^{2} = {u}^{2} + 2as \\ \\ \implies ({7.5})^{2} = ({2.5})^{2} + 2 \times 0.5 \times s \\ \\ \implies 56.25 = 6.25 + s \\ \\ \implies s = 56.25 - 6.25 \\ \\ \bold{\implies s = 50 \: m}$

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