Question

The velocity of a car increases from 30 kmph to 45 kmph in 5 minutes. Assuming the acceleration to be uniform ,calculate the distance travelled?

Answer 1

hope it helps you...

Answer 2

Answer:

The distance traveled during acceleration in 5 minutes is 3.12459 Km .

Explanation:

Provided that:

Initial velocity (u) = 30 Km/hour = 8.333 m/sec

Final velocity (v) = 45 Km/h = 12.5 m/sec

Time taken by car in this acceleration

(t) = 5 minutes = 300 seconds

Let, the acceleration is a m/sec^2

Calculating the acceleration we get;

$a = \frac{(v - u)}{t} \\ a = \frac{(12.5 - 8.333)}{300} \: m . {sec}^{ - 2} \\ a = 0.01389\: m . {sec}^{ - 2}$

Let the distance traveled during this acceleration is S ;

From Laws of Kinematics we know that;

${v}^{2} = {u}^{2} + 2 \: a \: S \\ or \: S = \frac{{v}^{2} - {u}^{2} }{2a} \\ or \: S = \frac{{(12.5)}^{2} - {(8.333)}^{2} }{(2 \times 0.01389)} \: m \\ = \frac{86.8111}{0.02778} \: m \\ = 3124.95 \: m \\ = 3.12495 \: km$

So we get the distance traveled during acceleration in 5 minutes is 3.12459 Km .