Question

A train traveling at 27km/h is accelerated at the rate of 0.5m/s. What is the distance traveled by the train in 12 seconds

Answer 1

Hey mate here is ur answer

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Explanation:

$given : \\initial \: velocity(u) \: = 27kmh {}^{ - 1} \\ = 27 \times \frac{5}{18} m{ s}^{ - 1} \\ = \frac{135}{18} m {s}^{ - 1} \\ = 7.5m {s}^{ - 1} \\ acceleration(a) = 0.5m {s}^{ - 2} \\ time \: taken \: (t) \: = 12s \\ distance(s) = {?} \\ solution: \\ s = ut + \frac{1}{2} at {}^{2} \\ s = (7.5) \times 12 + \frac{1}{2} \times 0.5 \times (12) {}^{2} \\ s = 90 + \frac{1}{2} \times 0.5 \times 144 \\ s = 90 + 0.5 \times 72 \\ s = 90 + 36 \\ s = 126m \: \: \: ans.$

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Answer 2

Given:-

Speed of train or initial velocity = 27 km/h

Acceleration of the train = 0.5 m/s²

Total time taken by train = 12 s

To Find:-

Distance travelled by train in 12s = ?

Solution:-

To calculate the distance travelled by train , at first we have to change the initial velocity into m/s then we use a formula to calculate the distance travelled by train. We use second equation of motion.

Formula Used:-

==> S = u t + 1/2 a t²

==> Initial velocity = 27 × 5/18

==> Initial velocity = 15/2 m/s

Let's Find out distance here:-

Let's Find out distance here:-As per 2nd equation of motion:-

==> S = Ut + 1/2 a t²

S = Distance , t = Time, U = initial velocity

==> S = 15/2 × 12 + 1/2 × 0.5 × (12)²

==> S = 15 × 6 + 1/2 × 0.5 × 12 × 12

==> S = 90 + 0.5 × 6 × 12

==> S = 90 + 0.5 × 72

==> S = 90 + 36

==> S = 126 m

Hence,

Distance travelled by train in 12s= 126 m