A car starts from rest and moves a distance of 5m in t seconds ,where s=1/6t³+1/4t².find the acceleration when t=0,t=2?
See the explanation below
Start by calculating the velocity, which is the derivative of the position
The acceleration is the derivative of the velocity
By signing up, you agree to our Terms of Service and Privacy Policy
To find the acceleration of the car when ( t = 0 ) and ( t = 2 ), we first need to differentiate the given displacement function ( s = \frac{1}{6}t^3 + \frac{1}{4}t^2 ) with respect to time ( t ) to obtain the velocity function, and then differentiate the velocity function to obtain the acceleration function.
Given: ( s = \frac{1}{6}t^3 + \frac{1}{4}t^2 )

Differentiate ( s ) with respect to ( t ) to find the velocity function ( v(t) ): [ v(t) = \frac{ds}{dt} = \frac{d}{dt}\left(\frac{1}{6}t^3 + \frac{1}{4}t^2\right) ]

Differentiate ( v(t) ) with respect to ( t ) to find the acceleration function ( a(t) ): [ a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2} = \frac{d^2}{dt^2}\left(\frac{1}{6}t^3 + \frac{1}{4}t^2\right) ]
Let's differentiate step by step:

( v(t) = \frac{d}{dt}\left(\frac{1}{6}t^3 + \frac{1}{4}t^2\right) ) [ v(t) = \frac{1}{2}t^2 + \frac{1}{2}t ]

( a(t) = \frac{d^2}{dt^2}\left(\frac{1}{6}t^3 + \frac{1}{4}t^2\right) ) [ a(t) = \frac{d}{dt}\left(\frac{1}{2}t^2 + \frac{1}{2}t\right) ] [ a(t) = t + \frac{1}{2} ]
Now, we have the acceleration function ( a(t) = t + \frac{1}{2} ).
To find the acceleration when ( t = 0 ) and ( t = 2 ), substitute these values into the acceleration function:

When ( t = 0 ): [ a(0) = 0 + \frac{1}{2} = \frac{1}{2} ]

When ( t = 2 ): [ a(2) = 2 + \frac{1}{2} = 2.5 ]
So, the acceleration when ( t = 0 ) is ( \frac{1}{2} ) and when ( t = 2 ) is ( 2.5 ).
By signing up, you agree to our Terms of Service and Privacy Policy
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
 If an object with uniform acceleration (or deceleration) has a speed of #12 m/s# at #t=0# and moves a total of 145 m by #t=15#, what was the object's rate of acceleration?
 An object's two dimensional velocity is given by #v(t) = ( t^2, tt^2sin(pi/3)t)#. What is the object's rate and direction of acceleration at #t=4 #?
 An object is thrown vertically from a height of #12 m# at # 19 ms^1#. How long will it take for the object to hit the ground?
 What is the average speed of an object that is still at #t=0# and accelerates at a rate of #a(t) = t^32t+2# from #t in [2, 3]#?
 Lauren's SUV was detected exceeding the posted speed limit of 60 kilometers per hour, how many kilometers per hour would she have been traveling over the limit if she had covered a distance of 10 kilometers in 5 minutes?
 98% accuracy study help
 Covers math, physics, chemistry, biology, and more
 Stepbystep, indepth guides
 Readily available 24/7