Answer:
Acceleration of the car is [tex]1.43\ m/s^2[/tex].
Explanation:
It is given that,
Initial speed of the car, u = 29 m/s
Finally it reaches a speed of, v = 34 m/s
Distance, d = 110 m
We need to find the acceleration of the car as it speed up. It can be calculated using third law of motion as :
[tex]v^2-u^2=2ad[/tex]
[tex]a=\dfrac{v^2-u^2}{2d}[/tex]
[tex]a=\dfrac{(34)^2-(29)^2}{2\times 110}[/tex]
[tex]a=1.43\ m/s^2[/tex]
So, the acceleration of the car as it speeds up is [tex]1.43\ m/s^2[/tex]. Hence, this is the required solution.
A car s traveling to the right with a speed of 29m/s when it accelerates at 1.4 m/s² for 110 m to reach a speed of 34 m/s.
A car is moving with an initial speed (u) of 29 m/s. After displacing for 110 m (s) with a constant acceleration (a), it reaches a speed of 34 m/s (final speed = v).
We can calculate the magnitude of the acceleration of the car using the following kinematic equation.
[tex]v^{2} = u^{2} + 2 \times a \times s\\\\a = \frac{v^{2}-u^{2}}{2 \times s} = \frac{(34m/s)^{2}-(29m/s)^{2}}{2 \times 110m} = 1.4 m/s^{2}[/tex]
A car s traveling to the right with a speed of 29m/s when it accelerates at 1.4 m/s² for 110 m to reach a speed of 34 m/s.
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