Question

Question 26:
A train starting from rest moves with a uniform acceleration of 0.2 m/s2 for 5 minutes. Calculate the speed acquired and the distance travelled in this time.

Lakhmir Singh Physics Class 9

Answer 1

acceleration,a = 0.2m/s2
time, t = 5 min = 300 s
initial velocity be u = 0m/s

using first equation of motion, final velocity:
v = u + at = 60 m/s

and according to second equation of motion, distance travelled,
s = ut + 1/2at2 = 9 km

Answer 2

Hi there !!
Here's your answer

Given that
A train is starting from rest
So,
initial speed (u) = 0m/s
Acceleration (a) = 0.2m/s²
Time (t) = 5min = 5 × 60 = 300 sec
Let the final speed be v

Using the 1st equation of motion
v = u + at

v = 0 + 0.2m/s² × 300sec

v = 60m/s

Thus,
the speed is 60m/s

To calculate the distance , we will use the 2nd equation of motion


$s = ut \: + \frac{1}{2} at {}^{2}$

$= 0 + \frac{1}{2} \times \frac{2}{10} \times 300 {}^{2}$

s = 9000 m

Thus,
the distance will be 9000 m or 9km

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Hope it helps :D