Question

5. A train starting from rest attains a velocity of 72 km per hour in 5 minutes. Assuming that the acceleration is uniform find : 1) acceleration and 2) the distance travelled by the train for attaining this velocity.

Answer 1

$\boxed{\boxed{\boxed{\purple{Hello \:Dear\:Students}}}}$


➖➖➖➖➖➖➖➖➖➖➖➖➖



Given :-

U = 0

T = 5 mins = 5×60 = 300 s

V = 72 km/h = 72 × $\frac{1000m}{60 \times 60s}$ = 20 m/s

Find : A = ? S = ?


(¡) acceleration = A = $\frac{V - U}{T}$


$\frac{20 - 0}{300}$


$\frac{1}{15}$ m/${s}^{2}$

(¡¡) distance

Use the Second Equation of Motion



S = UT + $\frac{1}{2}$ A${T}^{2}$



S = 0×300 $\frac{1}{2} \times \frac{1}{15}$ ${300}^{2}$



S = 3000 m



S = 3 km Ans



➖➖➖➖➖➖➖➖➖➖➖➖➖



Hope this helps . . . . . . .

Answer 2

Answer:

3km is the correct answer