Explanation:
Given,
Initial velocity of the car (u) = 10m/s
Acceleration of the car (a) = 2.5m/s²
Final velocity of the car (v) = 40m/s
(i) Let,
Time taken by the car to reach the final velocity = t
So,
By the problem,
=> v = u + at (According to first equation of motion)
=> 40 = 10 + 2.5t
=> 40 - 10 = 10 + 2.5t - 10
=> 30 = 2.5t
[tex] = > \frac{30}{2.5} = \frac{2.5t}{2.5} [/tex]
=> t = 12
Hence,
Required time taken by the car to reach a speed of 40m/s is 12 seconds. (Ans)
(ii) Let,
Distance traveled by the car within this time be = s
So,
By the problem,
=> s = ut + 1/2at² (According to second equation of motion)
[tex] = > s = 10 \times 12 + \frac{1}{2} \times 2.5 \times {12 }^{2} [/tex]
[tex] = > s = 10 \times 12 + \frac{1}{2} \times 2.5 \times 144 [/tex]
=> s = 120 + 180
=> s = 300
Hence,
Required distance covered by the car within that time is 300m (Ans)
