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Answer:
the acceleration is a=3ms^-2
Step-by-step explanation:
The distance is =69.3m
Apply the equation of motion
v2=u2+2as
the initial velocity is u=5ms−1
The final velocity is v=21ms−1
The acceleration is a=3ms−2
The distance is
s=v2−u22a
=212−522⋅3
=26×166
The average acceleration of the car is 4 m/s²
The given parameters:
initial velocity of the car, u = 2 m/s
final velocity of the car, v = 16 m/s
time of motion of the car, t = 3.5 s
To find:
- the average acceleration of the car
The average acceleration of the car is calculated as the change in velocity per change in time.
The formula for average acceleration is given below;
[tex]a = \frac{\Delta v}{\Delta t} = \frac{v- u}{t} = \frac{16 - 2}{3.5} = 4 \ m/s^2[/tex]
Thus, the average acceleration of the car is 4 m/s²
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