Question

a car starting from rest, accelerated uniformly at the rate of 10m/s^2. calculate (a) distance travelled in 5sec, (b) velocity at the end of 10sec (c) distance travelled in the 10th sec​

Answer 1

Answer:

a] Distance travelled in 5 sec = 125m.

b] velocity at end of 10 sec = 50 m/s.

c] distance travelled in 10th sec = 95m.

Explanation:

u = 0 m/s a = 10 m/s² t = 5sec

a] s = ut + 1/2 at²

s = 0×5 + 1/2 × 10 ×5 ×5

s = 5×5×5

s = 125m

b] v = u + at

v = 0 + 10 × 5

v = 50m/s

c] Formula for distance travelled in nth sec,

sₙ = u+1/2a(2n−1)

s₁₀ = 0 + 1/2 × 10 (2 ×10 -1)

s₁₀ = 5 × 19

s₁₀ = 95m

Answer 2

Answer:

• Initial velocity (u) = 0 m/s
• Acceleration (a) =10 m/s²
• Time elapsed (t) = 5 sec

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CALCULATION :

By using second kinematical equation of motion we can calculate distance travelled in 5sec :

$\longrightarrow s = ut + \dfrac{1}{2} {at}^{2}$

$\longrightarrow s = 0 \times 5 + \dfrac{1}{2} \times 10 \times {(5)}^{2}$

$\longrightarrow s = \dfrac{10}{2} \times 25$

$\longrightarrow s = 5 \times 25$

$\longrightarrow \underline{ \underline{ s = 1 25 \: m}}$

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By using Third kinematical equation of motion we can calculate the Final Velocity (v) :

$\longrightarrow {v}^{2} - {u}^{2} = 2as$

$\longrightarrow {v}^{2} - {(0)}^{2} = 2 \times 10 \times 125$

$\longrightarrow {v}^{2} = 20 \times 125$

$\longrightarrow {v}^{2} = 2500$

$\longrightarrow {v}^{2} = \sqrt{2500}$

$\longrightarrow \underline{ \underline{ v = 50 \: {ms}^{ - 1} }}$

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Distance travelled in the 10th seconds :-

$\longrightarrow S_n= u + \dfrac{a}{2} (2n - 1)$

$\longrightarrow S_{10}= 0 + \dfrac{10}{2} (2 \times 10- 1)$

$\longrightarrow S_{10}= 5(20- 1)$

$\longrightarrow S_{10}= 5(19)$

$\longrightarrow \underline{ \underline{ S_{10}= 95 \: m}}$

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