A car moving 30m/s slows uniformly to a speed of 10 m/s in a time of 5 sec determine

1) the acceleration of the car
2) the distance it moved in 3 sec

v=v₀-at

a= (v₀-v)/t
s=v₀t -at²/2

To find the answers, we will use the three equations of motion that relate acceleration, initial velocity, final velocity, and time.

1) The acceleration of the car can be determined using the equation:

acceleration = (final velocity - initial velocity) / time

Given:
Initial velocity (u) = 30 m/s
Final velocity (v) = 10 m/s
Time (t) = 5 sec

Using the equation, we can plug in the values:

acceleration = (10 m/s - 30 m/s) / 5 sec
acceleration = -20 m/s / 5 sec
acceleration = -4 m/s²

Therefore, the acceleration of the car is -4 m/s² (negative sign indicates deceleration).

2) To determine the distance the car moved in 3 seconds, we can use the equation:

distance = initial velocity * time + (1/2) * acceleration * time^2

Given:
Initial velocity (u) = 30 m/s
Time (t) = 3 sec

Using the equation, we can plug in the values:

distance = 30 m/s * 3 sec + (1/2) * (-4 m/s²) * (3 sec)^2
distance = 90 m + (1/2) * (-4 m/s²) * 9 sec^2
distance = 90 m - 18 m
distance = 72 m

Therefore, the car moved a distance of 72 meters in 3 seconds.