Answer :
(a) The acceleration of the car is,
![4.5m/s^2](https://img.qammunity.org/2020/formulas/physics/college/hqhcjnjn0rpsn3sjfwqxnzjj7sbay000e9.png)
(b) The distance covered by the car is, 9 m
Explanation :
By the 1st equation of motion,
...........(1)
where,
v = final velocity = 9 m/s
u = initial velocity = 0 m/s
t = time = 2 s
a = acceleration of the car = ?
Now put all the given values in the above equation 1, we get:
![9m/s=0m/s+a* (2s)](https://img.qammunity.org/2020/formulas/physics/college/h7q06i0vbnbdkq33pt21ewifazzs3trgha.png)
![a=4.5m/s^2](https://img.qammunity.org/2020/formulas/physics/college/1k47hhrq46elm1a52mhhur9r5qmlkne64x.png)
The acceleration of the car is,
![4.5m/s^2](https://img.qammunity.org/2020/formulas/physics/college/hqhcjnjn0rpsn3sjfwqxnzjj7sbay000e9.png)
By the 2nd equation of motion,
...........(2)
where,
s = distance covered by the car = ?
u = initial velocity = 0 m/s
t = time = 2 s
a = acceleration of the car =
![4.5m/s^2](https://img.qammunity.org/2020/formulas/physics/college/hqhcjnjn0rpsn3sjfwqxnzjj7sbay000e9.png)
Now put all the given values in the above equation 2, we get:
![s=(0m/s)* (2s)+(1)/(2)* (4.5m/s^2)* (2s)^2](https://img.qammunity.org/2020/formulas/physics/college/gbohvt014jri4lagepug0t68pawbd2q72y.png)
By solving the term, we get:
![s=9m](https://img.qammunity.org/2020/formulas/physics/college/gx3bcfb71c8mfqytpe7xey0dqsm8j46770.png)
The distance covered by the car is, 9 m