Question

A car initially at rest starts moving with a constant acceleration of 0.5ms^-2 and travels distance of 25m find final velociy and time taken

Answer 1

Initial velocity (u) = 0

Final velocity(v) = ?

Acceleration (a) = 0.5m/s²

Distance travelled (s) = 25

So, by third equation of motion,

v² = u² + 2as

v² = 0 + 2× 0.5 × 25

v² = 25

v² = (5)²

v = 5m/s²

To find time,

v = u + at

5 = 0 + 0.5t

5 = 0.5t

(1/2)t = 5

t = 5×2

t = 10seconds

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Answer 2

Given Initial velocity u = 0

Acceleration a = $\rm 0.5m {s}^{ - 2}$

Distance travelled S = 25m

i)Final velocity

From equation of motion $\rm {v}^{2}={u}^{2}+2aS$

$\implies \rm \: {v}^{2} = {0}^{2} + 2 \times 0.5 \times 25$

$\implies \rm \: {v}^{2} = 25$

$\implies \rm \: {v} = \sqrt{25}$

$\rm \fbox{ \therefore \: final \: velocity \: v \: = 5 ms{}^{ - 1} }$

ii) Time taken

From equation of motion v = u + at

$\implies$5 = 0 + 0.5 × t

$\implies$ 0.5 × t = 5

$\implies\rm{t = \frac{5}{0.5} }$

$\rm \fbox{\therefore \: time \: taken \: = 10s}$