Question

A body moving with an initial velocity of of 1 m/s accelerate uniformly at 0.5 m/s^2 find its displacement and velocity e at the end of 10 second what is the the velocity when it it has a a displacement of of 48 metre

Answer 1

GIVEN:

• Initial velocity, u = 1 m/s
• Acceleration, a = 0.5 m/s²

TO FIND:

• Displacement, s at the end of 10 seconds.
• Velocity, v at the end of 10 seconds.
• Velocity , v at a displacement of 48 m.

FORMULAE:

• According to the first equation of motion, v = u + at

• According to the second equation of motion, s = ut + (at²)/2

• According to the third equation of motion, v²- u² = 2as

SOLUTION:

STEP 1: To find displacement at the end of 10 seconds.

Let the displacement be s

u = 1 m/s

t = 10 s

a = 0.5 m/s²

$s = ut + \frac{1}{2} a {t}^{2} \\ \\ s = (1 \times 10) + ( \frac{1}{2} \times 0.5 \times {(10)}^{2} ) \\ \\ s = 10 + ( \frac{1}{2} \times 0.5 \times 10 \times 10 ) \\ \\ = 10 + ( \frac{50}{2} ) \\ \\ = 10 + 25 \\ \\ s = 35 \: m$

DISPLACEMENT AT THE END OF 10 SECONDS IS 35 m

STEP 2: To find velocity at the end of 10 seconds.

Let v be the final velocity.

u = 1 m/s

t = 10 s

a = 0.5 m/s²

$v = u + at \\ \\ v = 1 + (0.5 \times 10) \\ \\ v = 1 + 5 \\ \\ v = 6 \: \frac{m}{s}$

VELOCITY AT THE END OF 10 SECONDS IS 6 m/s

STEL 3: To find velocity when it has displaced 48 m.

s = 48 m

u = 1 m/s

a = 0.5 m/s²

${v}^{2} - {u}^{2} = 2as \\ \\ {v}^{2} - {(1)}^{2} = 2 \times 0.5 \times 48 \\ \\ {v}^{2} - 1 = 1.0 \times 48 \\ \\ {v}^{2} = 48 + 1 \\ \\ {v}^{2} = 49 \\ \\ v = \sqrt{49} \\ \\ v = 7 \: \frac{m}{s}$

VELOCITY WHEN IT HAS A DISPLACEMENT OF 48 m is 7 m/s.

ANSWERS:

• DISPLACEMENT AT THE END OF 10 SECONDS IS 35 m

• VELOCITY AT THE END OF 10 SECONDS IS 6 m/s

• VELOCITY WHEN IT HAS A DISPLACEMENT OF 48 m is 7 m/s.