Question

2 If the velocity of a train
starting from dest becomes
144kmhr in 5
minutes
a) what is its acceleration?
b) calculate the distance travelled
by the bain within this
time interval​

Answer 1

u=0

v=144km/h or 40m/s

t=5min or 5*60=300sec

a=v-u/t

a=40/300=0.133m/s²

s=v²-u²/2a=1600-0/2*0.133=6015m (approx)

Answer 2

Given

If the velocity of a train  starting from rest becomes  144 km/hr in 5  minutes

To Find

a. Acceleration

b. Distance traveled

Concept Used

We need to apply equations of kinematics .

→ v = u + at

→ s = ut + ¹/₂ at²

→ v² - u² = 2as

where ,

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time
• s denotes displacement

Solution

• u = 0 m/s

∵ starts from rest

• v = 144 km/h = 40 m/s
• t = 5 min = 300 s

Apply 1st equation of motion ,

⇒ v = u + at

⇒ 40 = 0 + a(300)

⇒ 300a  = 40

⇒ a = ²/₁₅

⇒ a = 0.134 m/s²  $\green{\bigstar}$

So , Acceleration = 0.134 m/s²

Apply 3rd equation of motion ,

⇒ v² - u² = 2as

⇒ (40)² - (0)² = 2(²/₁₅)s

⇒ 1600 - 0 = ⁴/₁₅ s

⇒ 4s = 1600 × 15

⇒ s = 400 × 15

⇒ s = 6000 m  $\pink{\bigstar}$

So , Distance traveled by the train = 6000 m