Question

2. A train starting from rest attains a velocity of 72 km/h in 5 minutes. Find (i) the acceleration.
(ii) the distance travelled by the train for attaining this velocity.
3. A car is traveling at 54 km/h. If brakes are applied so as to produce uniform retardation of 5m/s2. Calculate the time taken by it to stop. (Hint: a= -5m/s2)
4. Find the initial velocity of a train, which is stopped in 25 seconds. The retardation produced in the train is 1.5 m/s2.

Answer 1

Answer:

Given that,

Acceleration a=−0.5m/s2

Speed v=90km/h=25m/s

Using equation of motion, 

v=u+at

Where, 

v = final velocity

u = initial velocity

a = acceleration

t = time

Put the value into the equation

Finally train will be rest so, final velocity,v=0

0=25−0.5t

25=0.5t

t=0.525

t=50 sec

Again, using equation of motion, 

S=ut+21at2

Where, s = distance

v = final velocity

u = initial velocity

a = acceleration

t = time

Put the value into the equation

Where S is distance travelled before stop

s=25×50−21×0.5×(50)2

s=625 m

So, the train will go before it is brought to rest is 625 m.

Hence, A is correct.

Answer 2

Answer 2.

Given; u = 0 m/s, v = 72 km/hr = 20 m/s and t = 5 min = 300 sec.

To find- acceleration (a) and distance (s).

First equation of motion:

v = u + at

Second equation of motion:

s = ut + 1/2 at²

Third equation of motion:

v² - u² = 2as

Using the First Equation Of Motion,

v = u + at

→ 20 = 0 + a(300)

→ 20 = 300a

→ 0.067 = a

ii) Using the Third Equation Of Motion,

v² - u² = 2as

→ (20)² - (0)² = 2(0.067)s

→ 400 = 1.34s

→ s = 298.51

Therefore, the acceleration of the train is 0.067 m/s² and distance covered is 298.51 m.

Answer 3.

Given; u = 54 km/hr = 15 m/s, v = 0 m/s, a = -5 m/s²

To find- time taken (t).

Using the First Equation Of Motion,

v = u + at

→ 0 = 15 + (-5)t

→ -15 = -5t

→ 15/5 = t

→ 3 = t

Therefore, the time taken by the car is 3 seconds.

Answer 4.

Given; t = 25 sec, a = -1.5 m/s² and v = 0 m/s.

To find- initial velocity (u).

Using the First Equation Of Motion,

v = u + at

→ 0 = u + (-1.5)(25)

→ -u = -37.5

→ u = 37.5

Therefore, the initial velocity of the train is 37.5 m/s.