Question

Acceleration of a car is 3m/s^(-2) for 6 seconds and -4.5m/s^(-2) for next 4 seconds. If initially the car was at rest, then find velocity and position at time t.

a) 18 m/s

b) 12 m/s

c) 15 m/s

d) 0 m/s

Answer 1

The final velocity of the car after both acceleration phases is 0 m/s, and the total displacement is 90 m.

Step-by-step explanation:

The question involves calculating the velocity of a car over a period of time with two different accelerations. First, we calculate the velocity at the end of the first acceleration phase:

Velocity after first phase (6 seconds at 3 m/s^2)
v = u + at
v = 0 m/s + (3 m/s2 * 6s)
v = 18 m/s

Now, we calculate the velocity after the car decelerates for 4 seconds at -4.5 m/s^2:

Velocity after second phase (4 seconds at -4.5 m/s^2)
v = u + at
v = 18 m/s + (-4.5 m/s^2 * 4s)
v = 18 m/s - 18 m/s
v = 0 m/s

To find the position, we'll add the displacement from both phases.

During the first phase:
S1 = ut + ½ at^2
S1 = (0 m/s * 6s) + ½ * (3 m/s^2) * (6s)^2
S1 = 54 m

During the second phase:
S2 = ut + ½ at^2
S2 = (18 m/s * 4s) + ½ * (-4.5 m/s^2) * (4s)^2
S2 = 72 m - 36 m
S2 = 36 m

The total displacement is the sum of S1 and S2:
S = S1 + S2
S = 54 m + 36 m
S = 90 m

The final velocity of the car is 0 m/s, and the total displacement is 90 m.