A car starts from rest and accelerates at a constant 1.5 m/s2. Some time later, it takes 2.1 s to cross over a 25-meter long bridge. How far did the car travel before getting onto the bridge?
The correct answer and explanation is :
The car traveled approximately 35.57 meters before reaching the bridge.
Explanation:
The problem involves kinematics with constant acceleration. The car starts from rest, meaning its initial velocity (uu) is 0. It accelerates at 1.5 m/s² until it reaches the bridge.
Step 1: Find the velocity at the moment the car reaches the bridge
While crossing the 25-meter bridge in 2.1 seconds, we use the kinematic equation: d=vinitial⋅t+12at2d = v_{\text{initial}} \cdot t + \frac{1}{2} a t^2
Plugging in the given values: 25=vinitial⋅(2.1)+12(1.5)(2.1)225 = v_{\text{initial}} \cdot (2.1) + \frac{1}{2} (1.5) (2.1)^2
Solving for vinitialv_{\text{initial}}, we get: vinitial=9.82 m/sv_{\text{initial}} = 9.82 \text{ m/s}
This is the velocity just before reaching the bridge.
Step 2: Determine the time taken to reach the bridge
Using the formula: v=u+atv = u + at
Since the car starts from rest (u=0u = 0): 9.82=(1.5)t9.82 = (1.5) t
Solving for tt, we get: t=6.55 secondst = 6.55 \text{ seconds}
So, it took 6.55 seconds to reach the bridge.
Step 3: Calculate the distance before the bridge
Using: d=ut+12at2d = ut + \frac{1}{2} a t^2 d=0+12(1.5)(6.55)2d = 0 + \frac{1}{2} (1.5) (6.55)^2 d=35.57 metersd = 35.57 \text{ meters}
Thus, the car traveled 35.57 meters before reaching the bridge.