A car starts from rest and accelerates uniformly along a straight horizontal road at 3 metres second squeared for 4 seconds,after which it continues with a constant velocity for a further 6 seconds.What is the magnitude of the maximum velocity of the car,i.e. The velocity of the car after accelerating for 4 seconds?PLEASE SHOW ALL THE CALCULATIONS

V = at = 3m/s^2 * 4s = 12m/s.

To find the magnitude of the maximum velocity of the car, we need to calculate the distance covered during the acceleration phase and then determine the velocity at the end of that phase.

1. Calculate the distance covered during the acceleration phase:
During uniform acceleration, we can use the formula: distance = initial velocity * time + (1/2) * acceleration * time^2.
Given:
Initial velocity (u) = 0 m/s (as the car starts from rest)
Acceleration (a) = 3 m/s^2
Time (t) = 4 seconds

Using the formula, we have:
distance = 0 * 4 + (1/2) * 3 * (4^2)
distance = 0 + (1/2) * 3 * 16
distance = 0 + 24
distance = 24 meters

2. Calculate the velocity at the end of the acceleration phase:
Velocity can be calculated using the formula: final velocity (v) = initial velocity + acceleration * time.
Given:
Initial velocity (u) = 0 m/s (as the car starts from rest)
Acceleration (a) = 3 m/s^2
Time (t) = 4 seconds

Using the formula, we have:
final velocity = 0 + 3 * 4
final velocity = 12 m/s

Therefore, after accelerating for 4 seconds, the magnitude of the maximum velocity of the car is 12 m/s.