Answer:
1) v = 21.2 m/s
2) S = 63.33 m
3) s = 61.257 m
4) Deceleration, a = -4.32 m/s²
Explanation:
1) Given,
The initial velocity of Inna, u = 6.8 m/s
The acceleration of Inna, a = 4.5 m/s²
The time of travel, t = 3.2 s
Using the first equation of motion, the final velocity is
v = u + at
= 6.8 + 4.5 x 3.2
= 21.2 m/s
The final velocity of Inna is, v = 21.2 m/s
2) Given,
The initial velocity of Lisa, u = 12 m/s
The final velocity of Lisa, v = 26 m/s
The acceleration of Lisa, a = 4.2 m/s²
Using the III equations of motion, the displacement is
v² = u² +2aS
S = (v² - u²) / 2a
= (26² -12²) / 2 x 4.2
= 63.33 m
The distance Lisa traveled, S = 63.33 m
3) Given,
The initial velocity of Ed, u = 38.2 m/s
The deceleration of Ed, d = - 8.6 m/s²
The time of travel, t = 2.1 s
Using the II equations of motion, the displacement is
s = ut + 1/2 at²
=38.2 x 2.1 + 0.5 x(-8.6) x 2.1²
= 61.257 m
Therefore, the distance traveled by Ed, s = 61.257 m
4) Given,
The initial velocity of the car, u = 24.2 m/s
The final velocity of the car, v = 11.9 m/s
The time taken by the car is, t = 2.85 s
Using the first equations of motion,
v = u + at
∴ a = (v - u) / t
= (11.9 - 24.2) / 2.85
= -4.32 m/s²
Hence, the deceleration of the car, a = = -4.32 m/s²
