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

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft