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Question
2 motion with constant acceleration (continued)
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- displacement given initial and final velocities and the constant acceleration of an object, what mathematical relationship would you use to find the displacement?
- acceleration a woman driving west along a straight road at a speed of 23 m/s sees a deer on the road ahead. she applies the brakes when she is 210 m from the deer. if the deer does not move and the car stops right before it hits the deer, what is the acceleration provided by the cars brakes?
- distance the airplane in figure 18 starts from rest and accelerates east at a constant 3.00 m/s² for 30.0 s before leaving the ground.
a. what was the planes displacement (δx)?
b. how fast was the airplane going when it took off?
Step1: Recall the kinematic - equation
The kinematic equation $v^{2}=v_{0}^{2}+2a\Delta x$ can be used to find displacement $\Delta x$ when initial velocity $v_{0}$, final velocity $v$, and acceleration $a$ are known. We can re - arrange it as $\Delta x=\frac{v^{2}-v_{0}^{2}}{2a}$.
Step2: Identify given values for problem 34
Initial velocity $v_{0}=23\ m/s$, final velocity $v = 0\ m/s$, and displacement $\Delta x=210\ m$.
Step3: Use the kinematic equation for acceleration
From $v^{2}=v_{0}^{2}+2a\Delta x$, we can solve for $a$. Rearranging gives $a=\frac{v^{2}-v_{0}^{2}}{2\Delta x}$. Substituting $v = 0$, $v_{0}=23\ m/s$, and $\Delta x = 210\ m$, we have $a=\frac{0-(23)^{2}}{2\times210}=\frac{- 529}{420}\approx - 1.26\ m/s^{2}$.
Step4: Identify values for problem 35a
Initial velocity $v_{0}=0\ m/s$, acceleration $a = 3.00\ m/s^{2}$, and time $t = 30.0\ s$. We use the equation $\Delta x=v_{0}t+\frac{1}{2}at^{2}$. Since $v_{0}=0$, $\Delta x=\frac{1}{2}\times3\times(30)^{2}=1350\ m$.
Step5: Identify values for problem 35b
We use the equation $v=v_{0}+at$. With $v_{0}=0\ m/s$, $a = 3.00\ m/s^{2}$, and $t = 30.0\ s$, we get $v=0 + 3\times30=90\ m/s$.
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- $\Delta x=\frac{v^{2}-v_{0}^{2}}{2a}$
- $a\approx - 1.26\ m/s^{2}$
- a. $\Delta x = 1350\ m$
b. $v = 90\ m/s$