79. A commuter train can minimize the time t between two stations by accelerating (a, = 0.1 m/s2) for a time
tį then undergoing a negative acceleration (az =
-0.5 m/s2) by using his brakes for a time ty. Since the
stations are only 1 km apart, the train never reaches its
maximum velocity. Find the minimum time of travelt,
and the time t.

Let the train remains under acceleration for time t₁ and under deceleration for time t₂ .

The final velocity under acceleration becomes the initial velocity of deceleration.

.1 t₁ = .5 t₂

t₁ = 5 t₂

Now , total distance travelled is 1000 m

1/2 a₁t₁² + 1/2 a₂ t₂² = 1000

a₁t₁² + a₂ t₂² = 2000

.1 x 25 t₂² + .5 t₂² = 2000

3 t₂² = 2000

t₂² = 666.67

t₂ = 25.82 s

t₁ = 25.82 x 5 = 129.1 s

Total time = 154.92 s

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.