Question

A particle starts from rest with the acceleration 6m/s2 which decreases to zero linearly with time in 10 seconds after which the car continues at a constant speed. Find the time required by the car to travel 400m from the start

Answer 1

Answer:

t0=50/3sec

Explanation:

The relationship between acceleration and time can be expressed as

a(t)=m⋅t+c .

Now at t=0 , a(0)=6

⟹c=6 .

And at t=10 , a(10)=0

⟹m=−0.6

⟹a(t)=−0.6t+6 for 0≤t≤10 sec and a(t)=0 ms−2 afterwards.

Since car is moving on straight line so magnitude of displacement is same as total distance traveled.

Y

Now velocity can be given by expression ,

v(t)=−0.3t2+6t for 0≤t≤10 sec and v(t)=30 ms−1 afterwards.

Also total displacement can be expressed as ,

s(t)=−0.1t3+3t2 for 0≤t≤10 sec and s(t)=200+30(t−10) m afterwards.

Now at t=10 sec , s=200 m .

⟹ The car will reaches the destination at some t>10 sec .

If the car reaches the destination at t=t0 , then

s(t0)=200+30(t0−10)=400

⟹t0=50/3 sec

Answer 2

Hola mate

Here is your answer -

a(t)=m⋅t+c .

Now at t=0 , a(0)=6

⟹c=6 .

And at t=10 , a(10)=0

⟹m=−0.6

⟹a(t)=−0.6t+6 for 0≤t≤10 sec and a(t)=0 ms−2 afterwards.

Since car is moving on straight line so magnitude of displacement is same as total distance traveled.

Now velocity can be given by expression ,

v(t)=−0.3t2+6t for 0≤t≤10 sec and v(t)=30 ms−1 afterwards.

Also total displacement can be expressed as ,

s(t)=−0.1t3+3t2 for 0≤t≤10 sec and s(t)=200+30(t−10) m afterwards.

Now at t=10 sec , s=200 m .

⟹ The car will reaches the destination at some t>10 sec .

If the car reaches the destination at t=t0 , then

s(t0)=200+30(t0−10)=400

⟹t0=50/3 sec(16.66sec)