Question

A body moving in a velocity of 100 meter per second after four second its velocity becomes 160meter persecond. Find its accerllation or distance

Answer 1

Solution :

⏭ Given:

✏ Initial velocity of body = 100mps

✏ Final velocity of body = 160mps

✏ Time interval = 4s

⏭ To Find:

• Acceleration
• Distance travelled

⏭ Concept:

✏ Since, it is uniform acceleratory motion first we have to calculate acceleration of the body after that we can find out the distance covered by body.

⏭ Calculation:

• Acceleration

$\mapsto \sf \: \red{v = u + at} \\ \\ \mapsto \sf \: 160 = 100 + 4a \\ \\ \mapsto \sf \: 60 = 4a \\ \\ \mapsto \: \boxed{ \sf{ \blue{a = 15 \: m {s}^{ - 2}}}} \: \bigstar$

• Distance travelled

$\mapsto \sf \: \green{ {v}^{2} - {u}^{2} = 2as} \\ \\ \mapsto \sf \: {160}^{2} - {100}^{2} = 2 \times 15 \times s \\ \\ \mapsto \sf \: 25600 - 10000 = 30s \\ \\ \mapsto \sf \: 15600 = 30s \\ \\ \mapsto \: \boxed{ \sf{ \pink{s = 520 \: m}}} \: \bigstar$

Answer 2

$\huge{ \mathtt{ \fbox{Solution :)}}}$

Given ,

Inital velocity (u) = 100 m/s

Final velocity (v) = 160 m/s

Time (t) = 4 sec

We know that , the rate of change of velocity is known as acceleration

$\large \mathtt{{ \fbox{Acceleration \: (a) = \frac{v - u}{t} }}}$

Thus ,

a = (160 - 100)/4

a = 60/4

a = 15 m/s²

Hence , the acceleration is 15 m/s²

We know that , the third equation of motion is given by

$\large \mathtt{ \fbox{ {(v)}^{2} - {(u)}^{2} = 2as}}$

Thus ,

(160)² - (100)² = 2 × 15 × S

25600 - 10000 = 30 × S

15600 = 30 × S

S = 15600/30

S = 520 m

Hence , the distance is 520 m

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