The total distance covered by the car is 80 + 180 + 75 = 335 meters.
How can you calculate the total distance covered by the car?
Initial speed (u) = 10 m/s
Acceleration (a) = 5 m/s²
Time (t) = 4 s
Final speed (v) = u + at = 10 + 5 * 4 = 30 m/s
Second Phase (Constant Speed):
Time (t) = 6 s
No change in speed, so final speed remains 30 m/s
Third Phase (Deceleration):
Initial speed (u) = 30 m/s
Deceleration (a) = -6 m/s² (negative for slowing down)
Time (t) is unknown. We need to find this.
Final speed (v) = 0 m/s (car stops)
Using the equation v = u + at, we can solve for t in the third phase:
0 = 30 - 6t
6t = 30
t = 5 seconds
Therefore, it takes the car 5 seconds to stop after decelerating.
b. V-t Graph:
The V-t graph is uploaded
Linearly increasing slope from 10 m/s to 30 m/s within 4 seconds (acceleration phase).
Horizontal line at 30 m/s for 6 seconds (constant speed phase).
Linearly decreasing slope from 30 m/s to 0 m/s within 5 seconds (deceleration phase).
c. Acceleration for Each Interval:
First Phase: 5 m/s² (acceleration)
Second Phase: 0 m/s² (constant speed, no acceleration)
Third Phase: -6 m/s² (deceleration)
d. a-t Graph:
The a-t graph is uploaded
Horizontal line at 5 m/s² (constant acceleration)
Horizontal line at 0 m/s² (constant speed, no acceleration)
Horizontal line at -6 m/s² (constant deceleration)
e. Total Distance Covered:
First Distance = u * t + 1/2 at² = 10 * 4 + 1/2 * 5 * 4² = 80 m
Second Distance = v * t = 30 * 6 = 180 m
Third Distance = v * t - 1/2 at² = 30 * 5 - 1/2 * 6 * 5² = 75 m
Therefore, the total distance covered by the car is 80 + 180 + 75 = 335 meters.
f. Distance Traveled in 4th and 12th Seconds:
4th Second: During the 4th second, the car covers a distance of v * t = 10 * 4 = 40 meters.
12th Second: During the 12th second, the car is already in the constant speed phase, so it covers a distance of v * t = 30 * 1 = 30 meters.