Question

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 1

Answer:

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

${\bold{(ii)Time=10\:sec}}$

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

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 :

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

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

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

Answer 2

Solution:

Given:

➜ 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.

Find:

(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?

Given:

➜ 2.5 m/s = initial velocity.

➜ 0.5 m/s² = acceleration.

➜ 2 seconds = Time.

Know terms:

➜ Initial velocity = (u)

➜ final velocity = (v)

➜ Acceleration = (a)

➜ Time = (t)

(1) Answer:

By using third equation of motion:

➜ s = ut + 1/2 at²

➜ s = 2.5 × 2 × 1/2 × 0.50 × 2²

➜ s = 5 + 0.25 × 4

➜ s = 5 + 1

➜ s = 6 cm

(2) Answer:

By using first equation of motion:

➜ v = u + at

➜ 7.5 = 2.5 + 0.5 × t

➜ 7.5 - 2.5 = 0.5t

➜ t = 5/10.5

➜ t = 10 seconds.

(3) Answer:

By suing second equation of motion:

➜ v² = u² + 2as

➜ (7.5)² = (2.5)² + 2 × 0.5 × 5

➜ 56.25 = 6.25 + 5

➜ s = 56.2 - 6.25

➜ s = 50 m

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