To find the acceleration of the car in the given scenario, we can divide it into two phases: the first phase when the car accelerates uniformly to reach a speed of 30 m/s, and the second phase when the car slows down uniformly to come to rest.
Phase 1: Acceleration phase Initial velocity (u) = 0 m/s (starting from rest) Final velocity (v) = 30 m/s Time (t) = 30 s
We can use the following kinematic equation to calculate the acceleration (a) during this phase:
v = u + at
Rearranging the equation to solve for acceleration:
a = (v - u) / t
Substituting the given values:
a = (30 m/s - 0 m/s) / 30 s a = 1 m/s²
Therefore, the acceleration during the first phase is 1 m/s².
Phase 2: Deceleration phase Initial velocity (u) = 30 m/s (maintaining a speed of 30 m/s) Final velocity (v) = 0 m/s (coming to rest) Time (t) = 40 s
Using the same kinematic equation as before:
a = (v - u) / t
Substituting the given values:
a = (0 m/s - 30 m/s) / 40 s a = -0.75 m/s²
Note that the negative sign indicates deceleration or acceleration in the opposite direction.
Therefore, the acceleration during the second phase is -0.75 m/s².
In summary, the car accelerates uniformly with an acceleration of 1 m/s² for 30 seconds and then decelerates uniformly with an acceleration of -0.75 m/s² for the next 40 seconds.