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A car moving with constant acceleration covers the distance between two points 180 m apart in 6sec. Its speed as it passes the second point is $45 m/s$ What is its acceleration and its speed at the first point

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- 5 {m/s}^{2}; 15 m/s

5 {m/s}^{2}; 15 m/s

- 15 {m/s}^{2}; 5 m/s

- 5 {m/s}^{2}; - 15 m/s

answer is D.

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Detailed Solution

s = 180 m,t = 6 sec,v = 45 m/s

$a = ?, u = ?$

$s = vt - \frac{1}{2} {at}^{2}$

$180 = (6) (45) - \frac{1}{2} a {(6)}^{2}$

$180 = 270 - 18 a$

$\Rightarrow 18 a = 270 - 180 = 90$

$\Rightarrow a = 5 {m/s}^{2}$

$a = 5$

$\frac{v - u}{t} = 5$

$\frac{45 - u}{6} = 5 \Rightarrow 45 - u = 30$

$\Rightarrow u = 15 m/s$

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