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QuestionA train stops and speeds up to 60 m/s60 \mathrm{~m/s} in 40 s40 \mathrm{~s}. Find its acceleration. A boat stops in 6 s6 \mathrm{~s} from 18 m/s18 \mathrm{~m/s}. Find its acceleration.

Studdy Solution

STEP 1

Assumptions for Problem121. The train starts from rest. . The final speed of the train is60 m/s.
3. The time taken to reach this speed is40 seconds.
4. The acceleration is constant.

STEP 2

The formula for acceleration is given by the change in velocity divided by the change in time.
a=ΔvΔta = \frac{\Delta v}{\Delta t}

STEP 3

Now, plug in the given values for the change in velocity and change in time to calculate the acceleration.
a=60 m/s40 sa = \frac{60 \, m/s}{40 \, s}

STEP 4

Calculate the acceleration.
a=60 m/s40 s=1. m/s2a = \frac{60 \, m/s}{40 \, s} =1. \, m/s^2The acceleration of the train is 1. m/s21. \, m/s^2.

STEP 5

Assumptions for Problem131. The initial speed of the boat is18 m/s.
2. The boat slows to a stop, so its final speed is0 m/s.
3. The time taken to stop is seconds.
4. The deceleration (negative acceleration) is constant.

STEP 6

The formula for acceleration is given by the change in velocity divided by the change in time.
a=ΔvΔta = \frac{\Delta v}{\Delta t}

STEP 7

Now, plug in the given values for the change in velocity and change in time to calculate the acceleration.
a=0 m/s−18 m/s6 sa = \frac{0 \, m/s -18 \, m/s}{6 \, s}

STEP 8

Calculate the acceleration.
a=−18 m/s6 s=−3 m/s2a = \frac{-18 \, m/s}{6 \, s} = -3 \, m/s^2The acceleration (actually, deceleration) of the boat is −3 m/s2-3 \, m/s^2.

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