Question #279861

A car running at a speed of 72 km/h is to be stopped in exactly 10 seconds. Find the uniform deceleration needed.

1
Expert's answer
2021-12-15T11:19:38-0500

Explanations & Calculations


  • The question mentions a uniform deceleration. For uniform acceleration/ deceleration 4 equations of motions can be used appropriately.

v=u+at,v2=u2+2as,s=ut+12at2,s=(v+u)t2\qquad\qquad \begin{aligned} \small v=u+at,\, v^2=u^2+2as, \,s=ut+\frac{1}{2}at^2,\, s=\frac{(v+u)t}{2} \end{aligned}

  • The appropriate equation to be used for this question is the simplest one which is v=u+at\small v=u+at as the initial speed (72kmh1\small 72kmh^{-1} ), final speed and the time are given.
  • To begin with the calculations, you need to convert the given initial speed into ms1\small ms^{-1} as the time is given in seconds.

72kmh1=72×1000m3600s=20ms1\qquad\qquad \begin{aligned} \small 72\,kmh^{-1}&=\small \frac{72\times1000\,m}{3600\,s}\\ &=\small 20\,ms^{-1} \end{aligned}

  • Now you can try substituting the values to the equation and getting the answer.
  • The answer should come with a negative sign as the finding is the deceleration, not the acceleration.

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